Trigonometric ratios beyond acute angles AI HL

IB Mathematics: Applications and Interpretation HL · Unit 3

Lesson 3.8 · Trigonometric ratios beyond acute angles

Use the unit circle, interpret sine and cosine graphs, and solve trigonometric problems within a specified interval.

Praise from sunrise to sunset

“From the rising of the sun unto the going down of the same the LORD’s name is to be praised.” Psalm 113:3 (KJV)

As the sun appears to travel across the sky, its direction changes while the call to praise God remains. Angles can also turn beyond a single quadrant or make several complete revolutions. Let each new day remind you to praise the Lord throughout all its changing moments.

Learning objectives

By the end of this lesson, you should be able to:

define sine, cosine, and tangent using a point on the unit circle;
use the identity $\cos^2\theta+\sin^2\theta=1$ and determine signs in different quadrants;
explain how rotation around the unit circle produces sine and cosine graphs;
recognize and solve the ambiguous case of the sine rule; and
solve trigonometric equations graphically within a stated interval.

1. Sine and cosine on the unit circle

The unit circle has centre $(0,0)$ and radius $1$ . Start at the positive horizontal axis and rotate through an angle $\theta$ . If the endpoint is $P(x,y)$ , then:

$P=(\cos\theta,\sin\theta)$

Thus, cosine gives the horizontal coordinate and sine gives the vertical coordinate. These definitions work for angles in every quadrant, including negative angles and angles greater than one full turn.

In quadrant II, the point lies left of the vertical axis and above the horizontal axis: $\cos\theta<0$ and $\sin\theta>0$ .

Quadrant	$\sin\theta$	$\cos\theta$	$\tan\theta$
I	Positive	Positive	Positive
II	Positive	Negative	Negative
III	Negative	Negative	Positive
IV	Negative	Positive	Negative

Worked example · An angle beyond one revolution

Determine the quadrant and signs of the trigonometric ratios for $410^\circ$ .

$410^\circ-360^\circ=50^\circ$

The endpoint is the same as for $50^\circ$ , in quadrant I. Therefore sine, cosine, and tangent are all positive.

2. The Pythagorean identity and tangent

Every point $(x,y)$ on the unit circle satisfies $x^2+y^2=1$ . Replacing $x$ and $y$ with cosine and sine gives:

$\cos^2\theta+\sin^2\theta=1$

When $\cos\theta\ne 0$ , tangent is defined by:

$\tan\theta=\frac{\sin\theta}{\cos\theta}$

At $90^\circ$ and $270^\circ$ , and at angles differing from these by complete turns, $\cos\theta=0$ , so tangent is undefined.

Worked example · Use the quadrant to choose a sign

Suppose $\sin\theta=\frac{3}{5}$ and $\theta$ lies in quadrant II. Find $\cos\theta$ and $\tan\theta$ .

$\cos^2\theta=1-\left(\frac35\right)^2=\frac{16}{25}$

Cosine is negative in quadrant II, so $\cos\theta=-\frac45$ . Consequently:

$\tan\theta=\frac{3/5}{-4/5}=-\frac34$

Calculator check: Match the calculator’s degree or radian setting to the angle in the question. The quadrant determines which sign to use after taking a square root.

3. From the unit circle to graphs

Imagine a point moving steadily around the unit circle while $\theta$ increases. Its vertical coordinate traces $y=\sin\theta$ ; its horizontal coordinate traces $y=\cos\theta$ . Both coordinates stay between $-1$ and $1$ . After one complete turn, the point returns to its starting position, so both graphs repeat every $360^\circ$ , or $2\pi$ radians.

$\sin(\theta+360^\circ)=\sin\theta,\qquad \cos(\theta+360^\circ)=\cos\theta$

For instance, as $\theta$ goes from $0^\circ$ to $360^\circ$ , the sine graph passes through $0,1,0,-1,0$ at quarter-turn intervals. The cosine graph passes through $1,0,-1,0,1$ .

Worked example · Read coordinates after a turn

Find $\sin 240^\circ$ and $\cos 240^\circ$ , using the reference angle $60^\circ$ .

The angle $240^\circ=180^\circ+60^\circ$ ends in quadrant III, where both coordinates are negative:

$\sin240^\circ=-\frac{\sqrt3}{2},\qquad \cos240^\circ=-\frac12$

These familiar exact values help explain the graphs, but the IB syllabus does not require memorizing exact trigonometric values for assessment.

4. The ambiguous case of the sine rule

The sine rule relates a side to its opposite angle. If $a$ is opposite $A$ and $b$ is opposite $B$ , then:

$\frac{\sin A}{a}=\frac{\sin B}{b}$

When two sides and an angle opposite one of them are known, calculating $\sin B$ may give two possible angles: $B_1$ and $180^\circ-B_1$ . Check whether each makes a triangle with the known angle. There may be two triangles, one triangle, or no triangle.

Worked example · Two possible triangles

In triangle $ABC$ , $A=35^\circ$ , $a=8$ cm, and $b=12$ cm. Find the possible values of $B$ and $C$ .

$\sin B=\frac{b\sin A}{a}=\frac{12\sin35^\circ}{8}\approx0.86036$

$B_1\approx59.4^\circ,\qquad B_2=180^\circ-B_1\approx120.6^\circ$

Both work because $A+B<180^\circ$ in each case. Therefore:

$C_1\approx85.6^\circ,\qquad C_2\approx24.4^\circ$

These are two different triangles consistent with the measurements. Keep extra calculator digits until the final rounding step.

Quick checks: If the calculated sine exceeds $1$ , there is no triangle. If the second candidate angle makes the angle sum at least $180^\circ$ , reject that candidate.

5. Graphically solving trigonometric equations

To solve $\sin x=k$ on a finite interval, graph $y=\sin x$ and $y=k$ . Every intersection within the stated interval is a solution. Use the same method for $\cos x=k$ or for intersections of two trigonometric graphs. State your calculator mode and inspect both endpoints.

Worked example · More than one intersection

Solve $\sin x=0.40$ for $0^\circ\le x\le720^\circ$ , giving answers to one decimal place.

Graph $y=\sin x$ and $y=0.40$ in degree mode. The intersections in the first revolution occur near $23.6^\circ$ and $156.4^\circ$ . The graph repeats $360^\circ$ later, giving:

$x\approx23.6^\circ,\ 156.4^\circ,\ 383.6^\circ,\ 516.4^\circ$

The endpoints $0^\circ$ and $720^\circ$ are not solutions.

Graphing window: For degree questions, set the horizontal bounds to the given interval and the vertical bounds around $-1.2$ to $1.2$ . For radian questions, use radian mode and enter the interval in radians.

IB-style Paper 1 practice

Give appropriate working and check the interval in every equation question.

A point on the unit circle corresponds to $\theta=135^\circ$ . State the signs of $\sin\theta$ , $\cos\theta$ , and $\tan\theta$ . [3 marks]

Show solution

$135^\circ$ is in quadrant II. Therefore $\sin\theta>0$ , $\cos\theta<0$ , and $\tan\theta<0$ .
Given $\cos\theta=-\frac{5}{13}$ and $90^\circ<\theta<180^\circ$ , find $\sin\theta$ and $\tan\theta$ . [4 marks]

Show solution

$\sin^2\theta=1-\frac{25}{169}=\frac{144}{169}$

Quadrant II gives $\sin\theta=\frac{12}{13}$ and $\tan\theta=-\frac{12}{5}$ .
The terminal side of an angle passes through $(-4,-3)$ . Find $\sin\theta$ , $\cos\theta$ , and $\tan\theta$ . [4 marks]

Show solution

The distance from the origin is $\sqrt{(-4)^2+(-3)^2}=5$ . Hence $\sin\theta=-\frac35$ , $\cos\theta=-\frac45$ , and $\tan\theta=\frac34$ .
Solve $\cos x=-0.6$ for $0^\circ\le x\le360^\circ$ , to one decimal place. [3 marks]

Show solution

The horizontal line $y=-0.6$ intersects $y=\cos x$ twice on the interval:

$x\approx126.9^\circ,\ 233.1^\circ$
In triangle $PQR$ , $P=40^\circ$ , $p=7.0$ cm, and $q=9.0$ cm. Find all possible values of $Q$ , to one decimal place. [4 marks]

Show solution

$\sin Q=\frac{9\sin40^\circ}{7}\approx0.82644$

The two candidates are $Q\approx55.7^\circ$ and $Q\approx124.3^\circ$ . Both give $P+Q<180^\circ$ , so both are possible.
Solve $\sin x=-0.5$ for $0\le x\le2\pi$ , giving answers to three decimal places. [3 marks]

Show solution

In radian mode, the graph intersects $y=-0.5$ in quadrants III and IV:

$x=\frac{7\pi}{6}\approx3.665,\qquad x=\frac{11\pi}{6}\approx5.760$

Advanced applications and modelling

Rotating wheel. A marker on a wheel of radius $0.8$ m starts at the rightmost point. The wheel turns anticlockwise through $210^\circ$ about an axle $1.5$ m above the ground. Find the marker’s horizontal displacement from the axle and its height above the ground. [5 marks]

Show solution

Scale the unit-circle coordinates by $0.8$ . The horizontal displacement is $0.8\cos210^\circ=-0.4\sqrt3\approx-0.693$ m: the marker is $0.693$ m left of the axle. Its height is $1.5+0.8\sin210^\circ=1.5-0.4=\boxed{1.10\text{ m}}$ .
Surveying ambiguity. A surveyor measures $A=30^\circ$ , $a=10$ km, and $b=15$ km in triangle $ABC$ . Find every possible value of $B$ and determine the corresponding values of side $c$ , to two decimal places. [7 marks]

Show solution

$\sin B=\frac{15\sin30^\circ}{10}=0.75$

Thus $B\approx48.59^\circ$ or $131.41^\circ$ . The corresponding values of $C$ are $101.41^\circ$ and $18.59^\circ$ . Using $\frac{c}{\sin C}=\frac{10}{\sin30^\circ}$ gives $c\approx19.60$ km or $c\approx6.38$ km, respectively.
Sinusoidal voltage. A voltage is modelled by $V(t)=12\sin(2\pi t)$ volts, where $t$ is measured in seconds. Find every time in $0\le t\le2$ when $V(t)=6$ V. [6 marks]

Show solution

Graph $y=12\sin(2\pi t)$ and $y=6$ over $0\le t\le2$ , or solve $\sin(2\pi t)=\frac12$ . There are two intersections per second:

$t=\frac1{12},\ \frac5{12},\ \frac{13}{12},\ \frac{17}{12}\text{ seconds}$

Approximately $0.0833$ , $0.4167$ , $1.0833$ , and $1.4167$ seconds.
Graph intersection. Use a graph to find all solutions of $\sin x=\cos x$ for $-\pi\le x\le2\pi$ . Give exact answers. [5 marks]

Show solution

Graph $y=\sin x$ and $y=\cos x$ over the stated interval. Their intersection values are:

$x=-\frac{3\pi}{4},\ \frac{\pi}{4},\ \frac{5\pi}{4}$

Substitution confirms that sine and cosine are equal at each angle.
Design constraint. In a proposed triangular frame, $A=50^\circ$ , $a=8$ cm, and $b=12$ cm. Determine whether a frame satisfying these measurements can be constructed. [4 marks]

Show solution

$\sin B=\frac{12\sin50^\circ}{8}\approx1.149$

No real angle has sine greater than $1$ . Therefore no triangle satisfies these measurements.

Reflection · Praise through every turn

A complete revolution returns to its starting point, yet the journey passes through many different positions. Psalm 113:3 calls us to praise the Lord from sunrise to sunset. Where might you pause during your day to give thanks to God, whether circumstances feel bright or difficult?

Lesson 3.8 · Trigonometric ratios beyond acute angles

Praise from sunrise to sunset

Learning objectives

1. Sine and cosine on the unit circle

2. The Pythagorean identity and tangent

3. From the unit circle to graphs

4. The ambiguous case of the sine rule

5. Graphically solving trigonometric equations

IB-style Paper 1 practice

Advanced applications and modelling

Reflection · Praise through every turn

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IB Mathematics: Applications and Interpretation HL 6 Sections80 LessonsLifetime

Curriculum

Trigonometric ratios beyond acute angles AI HL

Lesson 3.8 · Trigonometric ratios beyond acute angles

Praise from sunrise to sunset

Learning objectives

1. Sine and cosine on the unit circle

2. The Pythagorean identity and tangent

3. From the unit circle to graphs

4. The ambiguous case of the sine rule

5. Graphically solving trigonometric equations

IB-style Paper 1 practice

Advanced applications and modelling

Reflection · Praise through every turn

Related

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alphy school publishers

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