The general definitions of the six trigonometric functions from a point on the terminal side, the unit circle and quadrantal angles, reference angles, the signs of the functions by quadrant, and the three-step procedure for evaluating any angle.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 13 — Trigonometric Ratios and Functions
Evaluate Trigonometric Functions of Any Angle
Objectives
Five outcomes. The same six functions, freed from the triangle.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.3 Evaluate Trigonometric Functions of Any Angle §13.3, pp. 866-871 — the lesson these objectives are drawn from
Warm-up
Lesson 13.1 defined sine as opposite over hypotenuse, which needs an acute angle in a right triangle.
Discussion prompt
Lesson 13.2 gave you angles of 240 degrees and negative 50. What could the sine of 240 degrees possibly mean, when no right triangle has such an angle?
Hint: Put the angle on the coordinate plane and look at a point on its terminal side.
Answer:
Take any point (x, y) on the terminal side and let r be its distance from the origin. Then define the sine to be y over r.
\[ \sin\theta = \frac{y}{r}, \quad r = \sqrt{x^2+y^2} \]
For an acute angle this is exactly the old definition, since x and y are then the legs and r the hypotenuse. But now x and y may be negative, so the definition survives into every quadrant.
Concept
Let theta be in standard position and let (x, y) be a point on its terminal side at distance r from the origin. The six functions are defined by the ratios of x, y and r. Because r is always positive and x and y carry signs, the quadrant determines the sign of every value.
unit circle — The circle x squared plus y squared equals 1, centred at the origin with radius 1. On it the sine of an angle is simply the y-coordinate and the cosine is the x-coordinate of the point where the terminal side crosses.
\[ \sin\theta = \frac{y}{r}, \; \cos\theta = \frac{x}{r}, \; \tan\theta = \frac{y}{x} \]
These are sometimes called the circular functions, because they can be read straight off a circle rather than out of a triangle.
Figure (svg): Two columns comparing the right-triangle definitions with the general ones
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.3 Evaluate Trigonometric Functions of Any Angle §13.3, pp. 866-867
Section
Section 1
Concept
Pick any point (x, y) on the terminal side, compute r as its distance from the origin, and form the six ratios. Any point on that side gives the same answers.
\[ r = \sqrt{x^2+y^2}, \quad r > 0 \]
The value r is a distance and is therefore always positive, so all the sign information lives in x and y.
Figure (svg): A point on the terminal side defining the six ratios for any angle
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.3 Evaluate Trigonometric Functions of Any Angle §13.3, pp. 866-866 — General Definitions of Trigonometric Functions
Picture it
Example 1: the point negative 4, 3.
Figure (svg): A point on the terminal side defining the six ratios for any angle
The distance r is 5, and every ratio follows. Two of the six are positive and four negative, all decided by the signs of x and y.
Worked example
Example 1.
\[ \text{The point } (-4,3) \text{ lies on the terminal side of } \theta. \text{ Evaluate all six functions.} \]
Find r
Why: Sixteen plus 9 is 25.
\[ r = 5 \]
Write sine and cosine
Why: Y over r and x over r.
\[ \frac{3}{5}\text{ and } -\frac{4}{5} \]
Write tangent
Why: Y over x.
\[ -\frac{3}{4} \]
Write the three reciprocals
Why: R over y, r over x, x over y.
\[ \frac{5}{3}, -\frac{5}{4}, -\frac{4}{3} \]
Figure (svg): A point on the terminal side defining the six ratios for any angle
\[ \sin\theta = \tfrac{3}{5}, \; \cos\theta = -\tfrac{4}{5}, \; \tan\theta = -\tfrac{3}{4} \]
Verify: check the signs against the quadrant
Why: The point has x negative and y positive, so it is in quadrant two — where sine is positive and cosine and tangent negative. All six values match that pattern, which is the fastest available check on a sign slip.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.3 Evaluate Trigonometric Functions of Any Angle §13.3, pp. 866-866
Fill the middle
Example 1.
Fill in the blanks
r = \sqrt5 = \sqrt___ = ___
Why: Five. Squaring negative 4 gives positive 16, which is why r comes out positive no matter which quadrant the point is in.
Worked example
Guided Practice 1 to 3.
\[ \text{Evaluate the six functions for the points } (3,-3), \; (-8,15) \text{ and } (-5,-12). \]
First: find r
Why: Nine plus 9 is 18.
\[ r = 3 \sqrt{2} \]
First: write the ratios
Why: In quadrant four.
\[ \sin - \sqrt{2} / 2, \cos \sqrt{2} / 2, \tan - 1 \]
Second: find r and the ratios
Why: Sixty-four plus 225 is 289.
\[ r = 17; \frac{15}{17}, -\frac{8}{17}, -\frac{15}{8} \]
Third: find r and the ratios
Why: Twenty-five plus 144 is 169.
\[ r = 13; - \frac{12}{13}, -\frac{5}{13}, \frac{12}{5} \]
Figure (svg): The solution to Worked example three more points shown as a ladder of expressions, one row per algebraic move
\[ r = 3\sqrt2; \quad r = 17; \quad r = 13 \]
Verify: notice the third point's positive tangent
Why: Both coordinates are negative, so their quotient is positive — and quadrant three is indeed where tangent is positive while sine and cosine are not. Two negatives dividing to a positive is the whole reason tangent behaves differently from the other two.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.3 Evaluate Trigonometric Functions of Any Angle §13.3, pp. 867-867
Trap
\[ (-4,3): \; r = -5? \]
Match the sign of r to the sign of x
Why: Since x is negative, r is made negative too.
\[ \sin\theta = \tfrac{3}{-5} = -\tfrac{3}{5} \quad \text{(wrong)} \]
The value r is a DISTANCE from the origin, and distances are never negative. Sine is positive in quadrant two.
\[ r = \sqrt{(-4)^2+3^2} = \sqrt{25} = 5 \]
Take the positive square root always
Why: The radical sign already means the positive root.
\[ \sin\theta = \tfrac{3}{5} > 0 \]
All the sign information belongs to x and y. Letting r carry a sign would double-count it and break every quadrant rule.
Matching
The signs of x and y decide everything.
Match the pairs
Why: Sine takes its sign from y alone, since r is always positive. The two quadrant-two points both have positive sines, and the two points with negative y both have negative ones.
Sorting
Read the signs of the coordinates.
Sort into buckets
Sort each point by the quadrant its terminal side lies in.
The last two differ only in the sign of x, which flips them between quadrants one and two and reverses the sign of the cosine while leaving the sine untouched.
Prediction
Commit before reasoning.
Predict first
The points (-4, 3) and (-8, 6) both lie on the same terminal side. Do they give the same sine?
Correct: Yes — r doubles to 10, and 6 over 10 is still 3 over 5.
\[ \frac{6}{10} = \frac{3}{5} \]
Why: Moving further out along the same ray scales x, y and r all by the same factor, so every ratio is unchanged. This is the same similarity argument that made the right-triangle ratios independent of triangle size, restated in coordinates. It is what makes these genuine functions of the ANGLE rather than of the chosen point.
Section
Section 2
Concept
The unit circle is x squared plus y squared equals 1. Since r is 1 there, sine is just the y-coordinate and cosine just the x-coordinate of the crossing point. Quadrantal angles, whose terminal sides lie on an axis, are read off it directly.
\[ \sin\theta = y, \quad \cos\theta = x \quad (r = 1) \]
Quadrantal angles are the multiples of 90 degrees, or pi over 2 radians, and are the only angles where a function can be undefined.
Figure (svg): The unit circle with its four quadrantal points
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.3 Evaluate Trigonometric Functions of Any Angle §13.3, pp. 867-867 — The Unit Circle
Picture it
The unit circle and its quadrantal crossings.
Figure (svg): The unit circle with its four quadrantal points
At 270 degrees the point is 0, negative 1, so the cosine is 0 — and any function with that 0 underneath is undefined.
Worked example
Example 2.
\[ \text{Use the unit circle to evaluate all six functions of } \theta = 270^\circ. \]
Find the crossing point
Why: Three quarters of a turn.
\[ (0, -1) \]
Read sine and cosine
Why: Y is negative 1 and x is 0.
\[ \sin = -1, \cos = 0 \]
Form the tangent
Why: Y over x is negative 1 over 0.
Form the reciprocals
Why: R over y, r over x, x over y.
Figure (svg): The unit circle with its four quadrantal points
\[ \sin 270^\circ = -1, \; \cos 270^\circ = 0, \; \tan 270^\circ \text{ undefined} \]
Verify: check which two are undefined
Why: Tangent and secant both have x underneath, and x is 0 here, so both fail together. Cotangent and cosecant have y underneath, and y is negative 1, so both are fine. Which pair breaks is decided entirely by which coordinate is zero.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.3 Evaluate Trigonometric Functions of Any Angle §13.3, pp. 867-867
Fill the middle
Example 2.
Fill in the blanks
\theta = 270^\circ \;\Longrightarrow\; (x,y) = (0,-1) \;\Longrightarrow\; \sin\theta = -1
Why: Negative 1. On the unit circle sine is simply the y-coordinate, which is as low as it can go — the minimum value sine ever takes.
Worked example
Guided Practice 4.
\[ \text{Use the unit circle to evaluate all six functions of } \theta = 180^\circ. \]
Find the crossing point
Why: Half a turn.
\[ (-1, 0) \]
Read sine and cosine
Why: Y is 0 and x is negative 1.
\[ \sin = 0, \cos = -1 \]
Form the tangent
Why: Zero over negative 1.
\[ \tan = 0 \]
Form the reciprocals
Why: Now y is the zero.
Figure (svg): The solution to Worked example the functions at 180 degrees shown as a ladder of expressions, one row per algebraic move
\[ \sin 180^\circ = 0, \; \cos 180^\circ = -1, \; \csc 180^\circ \text{ undefined} \]
Verify: compare with the 270 degree case
Why: There the zero was in x and tangent and secant failed; here the zero is in y and cosecant and cotangent fail instead. The pattern is exact: a function is undefined precisely when the coordinate in its denominator is zero.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.3 Evaluate Trigonometric Functions of Any Angle §13.3, pp. 867-867
Error analysis
A student evaluates the tangent of 270 degrees.
Annotate
On: \( \tan 270^\circ = \frac{-1}{0} = 0 \)
Zero on top gives zero; zero underneath gives undefined. Confusing the two positions is the commonest error with quadrantal angles.
Matching
Where the terminal side crosses the unit circle.
Match the pairs
Why: Reading down the right column gives the sines 0, 1, 0, -1 and the cosines 1, 0, -1, 0. Those eight numbers are the skeleton of the sine and cosine graphs that Lesson 14.1 will draw.
Sorting
Check the denominator's coordinate.
Sort into buckets
Sort each value.
Tangent and secant fail wherever x is zero, at 90 and 270 degrees. Cotangent and cosecant fail wherever y is zero, at 0 and 180. Sine and cosine, with r underneath, never fail at all.
Prediction
Commit before reasoning.
Predict first
Sine and cosine are defined for every angle whatsoever. Why?
Correct: Because their denominator is r, which is a distance and is never zero for a point on a terminal side.
\[ r > 0 \text{ always} \;\Longrightarrow\; \sin\theta, \cos\theta \text{ always defined} \]
Why: Any point on a terminal side other than the origin itself has r strictly positive, so y over r and x over r always make sense. The other four functions have x or y underneath, and those genuinely do hit zero. This is why the graphs of sine and cosine in Chapter 14 are unbroken waves while the tangent graph has vertical asymptotes.
Section
Section 3
Concept
For any non-quadrantal angle, the reference angle is the acute angle formed by the terminal side and the x-axis. It is found by subtracting from or subtracting off 180 or 360 depending on the quadrant.
\[ \text{II: }180^\circ-\theta; \; \text{III: }\theta-180^\circ; \; \text{IV: }360^\circ-\theta \]
The measurement is always to the x-axis, never to the y-axis, which is why the reference angle of a quadrant-two angle is 180 minus it rather than the angle minus 90.
Figure (svg): The reference angle in each of the three non-first quadrants
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.3 Evaluate Trigonometric Functions of Any Angle §13.3, pp. 868-868 — Reference Angle Relationships
Picture it
The reference angle in quadrants two, three and four.
Figure (svg): The reference angle in each of the three non-first quadrants
In every case it is the shortest swing back to the horizontal axis, so it always comes out between 0 and 90 degrees.
Worked example
Example 3, both parts.
\[ \text{Find the reference angle for } \theta = \tfrac{5\pi}{3} \text{ and for } \theta = -130^\circ. \]
First: locate the quadrant
Why: Five pi over 3 is 300 degrees.
First: apply the formula
Why: Two pi minus 5 pi over 3.
\[ \frac{\pi}{3} \]
Second: find a coterminal angle
Why: Negative 130 plus 360.
\[ 230 ^\circ,\text{ quadrant three} \]
Second: apply the formula
Why: Two thirty minus 180.
\[ 50 ^\circ \]
Figure (svg): The reference angle in each of the three non-first quadrants
\[ \theta' = \tfrac{\pi}{3}; \qquad \theta' = 50^\circ \]
Verify: check both are acute
Why: Pi over 3 is 60 degrees and 50 degrees is 50 degrees, both safely between 0 and 90. A reference angle outside that range means the wrong quadrant formula was used, which is the single most useful check here.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.3 Evaluate Trigonometric Functions of Any Angle §13.3, pp. 868-868
Fill the middle
Example 3a.
Fill in the blanks
\theta = \tfrac1___ \;\Longrightarrow\; \theta' = 2\pi-\tfrac______ = \tfrac___\pi}___
Why: Pi over 3, which is 60 degrees. Writing 2 pi as 6 pi over 3 makes the subtraction immediate.
Worked example
Guided Practice 5 to 8.
\[ \text{Find the reference angle for } 210^\circ, \; -260^\circ, \; -\tfrac{7\pi}{9}, \; \tfrac{15\pi}{4}. \]
First: 210 degrees
Why: Quadrant three; subtract 180.
\[ 30 ^\circ \]
Second: -260 degrees
Why: Add 360 to get 100, quadrant two.
\[ 180 - 100 = 80 ^\circ \]
Third: -7 pi over 9
Why: Add 2 pi to get 11 pi over 9, quadrant three.
\[ 11 \pi / 9 - \pi = 2 \pi / 9 \]
Fourth: 15 pi over 4
Why: Subtract 2 pi to get 7 pi over 4, quadrant four.
\[ 2 \pi - 7 \pi / 4 = \frac{\pi}{4} \]
Figure (svg): The solution to Worked example four more shown as a ladder of expressions, one row per algebraic move
\[ 30^\circ; \; 80^\circ; \; \tfrac{2\pi}{9}; \; \tfrac{\pi}{4} \]
Verify: check the two radian answers are acute
Why: Two pi over 9 is 40 degrees and pi over 4 is 45 degrees, both acute. Notice that both negative and oversized angles needed a coterminal reduction first — the quadrant formulas assume a measure between 0 and 360.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.3 Evaluate Trigonometric Functions of Any Angle §13.3, pp. 869-869
Trap
\[ \theta = 120^\circ \text{, in quadrant two} \]
Measure back to the nearest axis
Why: The y-axis at 90 degrees is closer than the x-axis at 180.
\[ \theta' = 120^\circ-90^\circ = 30^\circ \quad \text{(wrong)} \]
The reference angle is defined against the X-AXIS specifically, however close the y-axis may be.
\[ \theta' = 180^\circ-120^\circ = 60^\circ \]
Always measure to the x-axis
Why: Quadrant two uses 180 minus theta.
\[ \tan 120^\circ = -\tan 60^\circ = -\sqrt3 \]
Using 30 would have given negative root 3 over 3, a completely different number. The x-axis is not a preference here; it is the definition.
Fill the middle
Example 3b.
Fill in the blanks
-130^\circ+360^\circ = 230^\circ \;\Longrightarrow\; \theta' = 230^\circ-180^\circ = 50^\circ
Why: Fifty degrees. The negative angle had to be made coterminal and positive before the quadrant formula could be applied.
Matching
Reduce first, then apply the quadrant formula.
Match the pairs
Why: Every reference angle came out acute, as they must. The first and last are special angles, so their function values are exactly known; the middle two need a calculator.
Prediction
Commit before reasoning.
Predict first
Why bother finding the reference angle at all?
Correct: Because the function value at theta equals the value at the acute reference angle, up to a sign the quadrant supplies.
\[ \tan 120^\circ = -\tan 60^\circ = -\sqrt3 \]
Why: The triangle formed by the terminal side and the x-axis is congruent to the acute triangle at the reference angle, so the ratios match in size. Only the signs of x and y differ, and the quadrant records those. That is why one table of acute values, plus a quadrant rule, evaluates every angle there is — an enormous saving over tabulating all four quadrants separately.
Section
Section 4
Concept
Sine and cosecant follow the sign of y, cosine and secant follow the sign of x, and tangent and cotangent follow their quotient. Reading off the signs of x and y in each quadrant produces the whole table.
\[ \text{QII: }\sin>0, \cos<0, \tan<0 \]
Only quadrant one has all six positive. Each other quadrant has exactly one pair positive and the other two pairs negative.
Figure (svg): The signs of the six functions in each quadrant
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.3 Evaluate Trigonometric Functions of Any Angle §13.3, pp. 868-868 — Signs of Function Values
Picture it
Which of the six functions are positive where.
Figure (svg): The signs of the six functions in each quadrant
Nothing here needs memorising. Sine has y on top, cosine has x, and tangent has y over x, so the coordinate signs do all the work.
Worked example
The table on page 868.
\[ \text{Explain why tangent is positive in quadrant three but sine and cosine are not.} \]
Note the coordinate signs
Why: In quadrant three both are negative.
\[ x < 0, y < 0 \]
Sine is y over r
Why: Negative over positive.
Cosine is x over r
Why: Negative over positive.
Tangent is y over x
Why: Negative over negative.
Figure (svg): The signs of the six functions in each quadrant
\[ \tan\theta = \frac{y}{x} = \frac{-}{-} > 0 \]
Verify: check with the point (-5, -12)
Why: Its tangent is negative 12 over negative 5, which is positive 12 over 5, while sine is negative 12 over 13 and cosine negative 5 over 13. The point confirms the reasoning exactly.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.3 Evaluate Trigonometric Functions of Any Angle §13.3, pp. 868-868
Sorting
Sine follows the sign of y.
Sort into buckets
Sort each quadrant by the sign of the sine there.
Sine is positive in the upper half plane and negative in the lower, which is exactly the statement that sine follows y. Cosine splits left from right the same way.
Worked example
The table on page 868.
\[ \text{State the sign of } \sin\theta, \cos\theta \text{ and } \tan\theta \text{ in each of the four quadrants.} \]
Quadrant one
Why: Both coordinates positive.
Quadrant two
Why: X negative, y positive.
\[ \sin +, \cos -, \tan - \]
Quadrant three
Why: Both negative.
\[ \sin -, \cos -, \tan + \]
Quadrant four
Why: X positive, y negative.
\[ \sin -, \cos +, \tan - \]
Figure (svg): The solution to Worked example sign in each quadrant shown as a ladder of expressions, one row per algebraic move
\[ \text{I: all}; \; \text{II: }\sin; \; \text{III: }\tan; \; \text{IV: }\cos \]
Verify: notice which function is positive where
Why: Reading counterclockwise from quadrant one the positive one is all, then sine, then tangent, then cosine. Each function is positive in exactly two adjacent quadrants, and each quadrant past the first has exactly one function positive with its reciprocal.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.3 Evaluate Trigonometric Functions of Any Angle §13.3, pp. 868-868
Trap
\[ \cos 210^\circ, \; \theta' = 30^\circ \]
Report the value at the reference angle
Why: The reference angle's cosine is positive, so the answer is taken as positive.
\[ \cos 210^\circ = \frac{\sqrt3}{2} \quad \text{(wrong)} \]
The reference angle supplies only the SIZE. Quadrant three has x negative, so the cosine there is negative.
\[ \theta' = 30^\circ, \; \text{quadrant III} \]
Take the size from the reference angle and the sign from the quadrant
Why: Two separate decisions.
\[ \cos 210^\circ = -\frac{\sqrt3}{2} \]
The reference angle is always acute, so it can never tell you a sign. That is the quadrant's job, always.
Matching
Which pair survives.
Match the pairs
Why: A function and its reciprocal always share a sign, since flipping a fraction never changes it. That is why the six functions come in three pairs and only three answers are ever needed.
Fill the middle
Example 4a.
Fill in the blanks
\tan(-240^\circ) = -\tan 60^\circ = -\sqrt3
Why: A minus sign. The angle lands in quadrant two where x is negative and y positive, so their quotient is negative.
Prediction
Commit before reasoning.
Predict first
Sine and cosecant are always positive together and negative together. Why must that be so?
Correct: Because cosecant is 1 over sine, and taking a reciprocal never changes a sign.
\[ \csc\theta = \frac{1}{\sin\theta} \;\Longrightarrow\; \text{same sign} \]
Why: One over a positive number is positive and one over a negative number is negative, so a reciprocal pair always agrees. That halves the work of learning the sign table: three facts about sine, cosine and tangent determine all six. The only place the pairing breaks down is where the primary function is zero, and there the reciprocal is undefined rather than differently signed.
Section
Section 5
Concept
To evaluate a trigonometric function at any angle: find the reference angle, evaluate the function there, and attach the sign that the original angle's quadrant demands.
\[ \tan(-240^\circ) = -\tan 60^\circ = -\sqrt3 \]
The first step usually needs a coterminal reduction, so the working order is reduce, locate, reference, evaluate, sign.
Figure (svg): The three-step evaluation applied to a negative angle
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.3 Evaluate Trigonometric Functions of Any Angle §13.3, pp. 868-869 — Evaluating Trigonometric Functions
Picture it
Example 4a: the tangent of negative 240 degrees.
Figure (svg): The three-step evaluation applied to a negative angle
The size comes from the acute angle 60 and the minus sign from quadrant two, and neither step can supply the other.
Worked example
Example 4, both parts.
\[ \text{Evaluate } \tan(-240^\circ) \text{ and } \csc\tfrac{17\pi}{6}. \]
First: reduce and locate
Why: Negative 240 plus 360 is 120.
First: reference angle and sign
Why: One eighty minus 120; tangent negative there.
\[ -\tan 60 = -\sqrt{3} \]
Second: reduce and locate
Why: Seventeen pi over 6 minus 2 pi is 5 pi over 6.
Second: reference angle and sign
Why: Pi minus 5 pi over 6; cosecant positive there.
\[ \csc(\frac{\pi}{6}) = 2 \]
Figure (svg): The three-step evaluation applied to a negative angle
\[ \tan(-240^\circ) = -\sqrt3; \qquad \csc\tfrac{17\pi}{6} = 2 \]
Verify: notice both angles are in the same quadrant with different signs
Why: Both land in quadrant two, yet the tangent is negative and the cosecant positive. That is the sign table doing exactly its job: quadrant two keeps sine and cosecant but reverses the other four.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.3 Evaluate Trigonometric Functions of Any Angle §13.3, pp. 869-869
Fill the middle
Example 4b.
Fill in the blanks
\tfrac5___-2\pi = \tfrac______-\tfrac______ = \tfrac___\pi}___
Why: Five pi over 6, which is 150 degrees and lies in quadrant two. Writing 2 pi with the same denominator makes the subtraction routine.
Worked example
Guided Practice 9, and Example 5.
\[ \text{Evaluate } \cos(-210^\circ). \text{ Then find how far a robot jumping at } 45^\circ \text{ at } 16 \text{ ft/s travels, using } d = \tfrac{v^2}{32}\sin 2\theta. \]
Reduce and locate
Why: Negative 210 plus 360 is 150.
Reference angle and sign
Why: One eighty minus 150; cosine negative there.
\[ -\cos 30 = -\sqrt{3}\text{ over } 2 \]
Substitute into the model
Why: Sixteen squared over 32, times sine of 90.
\[ 8 \times \sin 90 \]
Evaluate
Why: Sine of 90 is 1.
\[ d = 8\text{ feet} \]
Figure (svg): The solution to Worked example an evaluation and an application shown as a ladder of expressions, one row per algebraic move
\[ \cos(-210^\circ) = -\frac{\sqrt3}{2}; \qquad d = 8 \text{ ft} \]
Verify: check why 45 degrees is the best launch angle
Why: The model doubles the angle before taking the sine, and sine peaks at 90 degrees — which happens exactly when theta is 45. Any other launch angle gives a sine below 1 and so a shorter jump, which is why 45 degrees is the classic answer to how to throw something furthest.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.3 Evaluate Trigonometric Functions of Any Angle §13.3, pp. 869-869
Error analysis
A student evaluates the cosecant of 17 pi over 6.
Annotate
On: \( \theta' = 2\pi-\tfrac{17\pi}{6} = -\tfrac{5\pi}{6} \)
The quadrant formulas assume a measure between 0 and 360 degrees, or 0 and 2 pi. Reduce before locating, always.
Fill the middle
Example 4b.
Fill in the blanks
\csc\tfrac2___ = \frac______ = \frac______ = ___
Why: Two. Cosecant is positive in quadrant two, so the answer keeps its sign and the cosecant of 17 pi over 6 is 2.
Ranking
Smallest first.
Put in order
Why: The values are about -1.732, about -0.866, 0, 2 and 8. The first two are negative because their angles land in quadrant two where tangent and cosine both fail to survive.
Prediction
Commit before reasoning.
Predict first
In d = (v squared over 32) sin 2 theta, which launch angle sends a projectile furthest?
Correct: Forty-five degrees, because it makes 2 theta equal 90 where the sine is largest.
\[ \sin 2\theta \text{ peaks at } 2\theta = 90^\circ \]
Why: Sine reaches its maximum of 1 at 90 degrees, and 2 theta is 90 exactly when theta is 45. Launching straight up gives 2 theta equal to 180, whose sine is 0 — the projectile comes straight back down and travels no horizontal distance at all. The model also predicts that 30 and 60 degrees give identical ranges, since sine of 60 and sine of 120 are equal, which is a genuinely surprising consequence that turns out to be true.
Comparison
Fill the blanks. Same functions, wider reach.
Comparison matrix
| Question | Lesson 13.1 | Lesson 13.3 |
|---|---|---|
| Definition of sine | opposite over hypotenuse | y over r |
| Angles allowed | acute only | any angle at all |
| Signs | always positive | set by the quadrant |
| Undefined values | none | at quadrantal angles |
In quadrant one the two definitions agree exactly, because there x, y and r are the adjacent leg, the opposite leg and the hypotenuse. Everything beyond quadrant one is genuinely new.
Pattern
Reduce, locate, reference, sign.
The reference angle gives only the size. Every sign comes from the quadrant.
Check
Find r first, and keep it positive.
Check your understanding
The point (-4, 3) is on the terminal side of theta. What is cos theta?
Answer: A
Why: r = 5 and cosine is x over r, so it is -4/5.
Check
Measure to the x-axis.
Check your understanding
What is the reference angle for 5 pi/3?
Answer: A
Why: 5 pi/3 is 300 degrees, in quadrant four, so the reference angle is 2 pi - 5 pi/3 = pi/3.
Check
Size from the reference angle, sign from the quadrant.
Check your understanding
What is tan(-240 degrees)?
Answer: A
Why: -240 is coterminal with 120, whose reference angle is 60; tangent is negative in quadrant two.
Real world
A Ferris wheel of radius 40 feet has its centre 45 feet above the ground. A rider boards at the lowest point and the wheel turns counterclockwise.
Discussion prompt
How high is the rider after the wheel has turned through 210 degrees?
Hint: Boarding at the bottom means starting at an angle of 270 degrees in standard position.
Answer:
\[ \text{start at } 270^\circ; \quad \text{after } 210^\circ: \; \theta = 480^\circ \equiv 120^\circ \]
\[ \text{height} = 45+40\sin 120^\circ = 45+40\!\left(\frac{\sqrt3}{2}\right) \]
\[ = 45+20\sqrt3 \approx 79.6 \text{ feet} \]
The rider is about 79.6 feet above the ground.
Two ideas from this lesson carry the whole calculation. The angle 480 degrees had to be reduced to its coterminal 120 before anything else could happen, and the sine of 120 needed the reference angle 60 together with the fact that sine stays positive in quadrant two. Notice also how naturally the height splits into a centre height plus 40 times a sine — that shape, a constant plus an amplitude times a sine, is exactly the model Lesson 14.4 will build for every periodic quantity.
Commit first
Answer, then rate your confidence honestly.
Predict first
Is the tangent of 270 degrees equal to zero, or undefined?
Correct: Undefined — tangent is y over x, and here x is 0, putting the zero underneath.
\[ \tan 270^\circ = \tfrac{-1}{0} \text{ undefined}; \quad \tan 180^\circ = \tfrac{0}{-1} = 0 \]
Why: The point on the unit circle is 0, negative 1, so tangent is negative 1 divided by 0, which has no value. The trap is that the coordinates do contain a zero, and a zero in a fraction sometimes does give zero — but only when it is on TOP. At 180 degrees the point is negative 1, 0 and the tangent is 0 over negative 1, which really is zero. Same two numbers, opposite positions, completely different outcomes.
Explain it
They know sine as opposite over hypotenuse and cannot see how a 200 degree angle has one.
Discussion prompt
In four sentences or fewer, explain how sine is defined for an obtuse angle.
Hint: Talk about a point rather than a triangle.
Answer:
Draw the angle starting from the positive x-axis and pick any point on the ray where it ends. Measure that point's distance from the origin and call it r.
The sine is the point's height divided by r. For an acute angle that is the old definition exactly, but now the height is allowed to be negative, so every angle has one.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For r, remember it is a distance and take the positive root always. For zeros, ask which position the zero occupies: on top gives zero, underneath gives undefined. For reference angles, always measure to the x-axis and check the answer is acute. For signs, decide them from the quadrant as a separate final step, never from the reference angle.
Connect it up
Paper. Fifteen minutes.
Draw it
Build an any-angle page. Top left: plot a point in each of the four quadrants, draw the terminal side and the r segment for each, and write all six ratios for one of them. Top right: draw the unit circle with its four quadrantal points labelled, and beside it list which two functions are undefined at each of 90, 180, 270 and 0, saying in each case which coordinate is the zero. Middle: draw the three reference-angle diagrams with their formulas, and beside each write one angle of your own and its reference angle. Bottom left: draw the four-quadrant sign chart, but derive it rather than copying it, writing the signs of x and y in each quadrant first. Bottom right: evaluate three functions at non-acute angles, showing all three steps separately and boxing the sign decision.
If any reference angle on your page came out above 90 degrees or negative, the quadrant formula was misapplied. Reference angles are always acute.
Recap
Five things, and six functions that now accept any angle.
| If you see | Then |
|---|---|
| A point on the terminal side | r is the positive root of x squared plus y squared |
| A multiple of 90 degrees | Read the point off the unit circle |
| A zero underneath | The function is undefined there |
| A non-acute angle | Reduce, locate, then find the reference angle |
| A reference angle value | It gives the size only, never the sign |
| Quadrant three | Only tangent and cotangent are positive |
Lesson 13.4 reverses the question: given a function value, which angle produced it? That is the inverse trigonometric function, and it needs care because many angles share a value.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.3 Evaluate Trigonometric Functions of Any Angle §13.3, pp. 866-871 — everything on these slides traces back here
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