Lesson 5 ~45 min Trigonometry D · Path +90 XP

Supplementary and Complementary Angles

Two pairs of relationships tie the whole circle together. Supplementary angles, adding to $180°$, share a sine and have opposite cosines. Complementary angles, adding to $90°$, exchange sine with cosine. Both follow from a single reflection.

Today's hook: The word cosine means the sine of the complement, and that is not a historical curiosity: it is the definition. Once the complementary relationship is established, half the identities in trigonometry stop needing to be remembered separately.

0/5QUESTS

Think First

warm-up

Mark the points at $50°$ and at $130°$ on a unit circle. The two angles add to $180°$. Compare the two points' heights and their horizontal positions, then say which of sine and cosine is unchanged and which has flipped.

Record your answer in your workbook.

The Big Idea

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Supplementary angles, adding to $180°$, give points that are mirror images in the vertical axis: same height, opposite horizontal position. Complementary angles, adding to $90°$, exchange the two coordinates entirely.

$$\sin(180° - \theta) = \sin\theta, \qquad \cos(90° - \theta) = \sin\theta$$

The complementary pair explains the name. Cosine is the sine of the complement: $\cos\theta = \sin(90° - \theta)$. In a right-angled triangle the two acute angles are complementary, and the side opposite one is the side adjacent to the other, which is the whole proof.

$\sin(180°-\theta) = \sin\theta$

Supplementary keeps sine

$180° - \theta$: sine survives, cosine and tangent change sign.

Complementary swaps

$90° - \theta$: sine and cosine exchange places.

Check with a known angle

Test any proposed identity at $30°$ or $45°$ before trusting it.

What You'll Master

objectives

Know

The supplementary relationships for sine, cosine and tangent
The complementary relationships between sine and cosine
That two supplementary angles share a sine, which is why some problems have two answers

Understand

Why the supplementary relationships follow from a reflection in the vertical axis
Why the complementary relationships follow from the two acute angles of a right triangle

Can Do

Establish each relationship from the unit circle or from a right triangle
Use them to rewrite a ratio of an obtuse angle in terms of an acute one
Find a second angle with a given sine

Words You Need

vocabulary

SupplementaryTwo angles summing to $180°$.

ComplementaryTwo angles summing to $90°$.

IdentityA relationship true for every value of the angle.

ReflectionA flip across a line. Reflecting in the vertical axis changes the sign of the first coordinate.

Co-ratioCosine is the sine of the complement, which is where its name comes from.

Supplementary Angles

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Take an angle $\theta$ and its supplement $180° - \theta$. Where do the two points sit on the unit circle?

Turning through $180° - \theta$ means turning almost half a turn, then coming back by $\theta$. The result is the mirror image, in the vertical axis, of the point at $\theta$: the same height, the opposite horizontal position.

Since sine is the height and cosine is the horizontal position:

$$\sin(180° - \theta) = \sin\theta$$

$$\cos(180° - \theta) = -\cos\theta$$

and dividing the first by the second gives

$$\tan(180° - \theta) = -\tan\theta$$

Check at $\theta = 30°$: the supplement is $150°$, and Lesson 3 gave $\sin 150° = \tfrac{1}{2} = \sin 30°$, with $\cos 150° = -\tfrac{\sqrt{3}}{2} = -\cos 30°$. Both agree.

The sine relationship is the important one, because it means two different angles between $0°$ and $180°$ share a sine. That is the reason a question asking for an angle from its sine can have two answers, and it is what the ambiguous case in Lesson 7 is about.

Complementary Angles

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Now take $\theta$ and its complement $90° - \theta$. The clearest proof here is the right triangle rather than the circle.

In a right-angled triangle the three angles sum to $180°$ and one is $90°$, so the other two sum to $90°$: they are complementary. Call them $\theta$ and $90° - \theta$.

Now look at the sides. The side opposite $\theta$ is the side adjacent to $90° - \theta$, and vice versa. The hypotenuse is shared. So:

$$\sin\theta = \frac{o}{h} \qquad \text{and} \qquad \cos(90° - \theta) = \frac{o}{h}$$

because the side that is opposite $\theta$ is exactly the side adjacent to the other angle. Therefore

$$\cos(90° - \theta) = \sin\theta \qquad \text{and} \qquad \sin(90° - \theta) = \cos\theta$$

Check with the special triangle: $\sin 30° = \tfrac{1}{2}$ and $\cos 60° = \tfrac{1}{2}$, and $30°$ and $60°$ are complementary. Also $\sin 45° = \cos 45°$, since $45°$ is its own complement.

Where the name comes from

"Cosine" is short for the sine of the complement, and "cotangent" and "cosecant" are named the same way. The relationship is not a fact about cosine so much as the reason the word exists.

Using Them

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The relationships have three standard uses.

Rewriting an obtuse ratio as an acute one. Any ratio of an angle between $90°$ and $180°$ can be converted:

$$\sin 137° = \sin(180° - 137°) = \sin 43°$$

$$\cos 137° = -\cos 43°$$

which is exactly the related-angle method of Lesson 3, now stated as an identity rather than a procedure.

Finding a second solution. If $\sin\theta = 0.6$ and one solution is $\theta \approx 36.87°$, then the supplement $180° - 36.87° = 143.13°$ is another, because supplementary angles share a sine. A calculator returns only the first, so the second must be found by you.

Simplifying an expression. Recognising a complementary pair can collapse an expression at once:

$$\frac{\sin 20°}{\cos 70°} = \frac{\sin 20°}{\sin 20°} = 1$$

since $70° = 90° - 20°$, so $\cos 70° = \sin 20°$.

Look for pairs summing to $90°$ or to $180°$ before doing any calculation. Spotting one often removes the calculation entirely.

Keeping the Two Pairs Apart

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The two sets of relationships are easy to confuse, and the differences are worth stating side by side.

Supplementary, $180° - \theta$. The ratio names stay the same; only signs change. Sine keeps its sign, cosine and tangent flip. Both angles are in the upper half of the circle, so both have positive sine.

Complementary, $90° - \theta$. The ratio names swap; no signs change, provided $\theta$ is acute so both angles are in the first quadrant. Sine becomes cosine and cosine becomes sine.

A one-line summary:

$$180° - \theta: \ \text{same name, maybe a minus} \qquad 90° - \theta: \ \text{swapped name, no minus}$$

The safest habit is to test any proposed identity at a known angle before using it. If you are unsure whether $\cos(180° - \theta)$ is $\cos\theta$ or $-\cos\theta$, put $\theta = 60°$: then $180° - 60° = 120°$, and $\cos 120° = -\tfrac{1}{2}$ while $\cos 60° = +\tfrac{1}{2}$. The minus sign is needed.

One test at one angle settles it in five seconds and is far more reliable than recalling which of four similar-looking statements is the right one.

Common Pitfalls

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Writing $\cos(180° - \theta) = \cos\theta$, without the minus sign.

Fix: test at $\theta = 60°$. $\cos 120° = -\tfrac{1}{2}$ and $\cos 60° = \tfrac{1}{2}$, so the sign must change.

Swapping sine and cosine for supplementary angles instead of complementary ones.

Fix: the swap belongs to $90° - \theta$. For $180° - \theta$ the names stay put and only signs move.

Giving only the calculator's answer when an angle is found from its sine.

Fix: the supplement is a second answer. If the context allows an obtuse angle, both must be given, which is exactly the ambiguous case of Lesson 7.

Applying the complementary relationship when the two angles do not actually sum to $90°$.

Fix: check the sum first. $\cos 70°$ equals $\sin 20°$ because $70 + 20 = 90$; it has no such simple relationship to $\sin 30°$.

Watch Me Solve It · 3 examples

Watch Me Solve It · Establishing the supplementary relationships

+15 XP per step

PROBLEM

Use the unit circle to establish $\sin(180° - \theta) = \sin\theta$ and $\cos(180° - \theta) = -\cos\theta$, then deduce the tangent relationship and verify all three at $\theta = 45°$.

1

Locate the second point

The point at $180° - \theta$ is the reflection, in the vertical axis, of the point at $\theta$. Reflecting in the vertical axis keeps the height unchanged and reverses the horizontal position.
2

Read off the two coordinates

$\sin(180° - \theta) = \sin\theta$

$\cos(180° - \theta) = -\cos\theta$

Sine is the second coordinate, which is unchanged; cosine is the first, which has changed sign.
3

Divide to get the tangent

$\tan(180° - \theta) = \frac{\sin\theta}{-\cos\theta} = -\tan\theta$

The tangent is the quotient, so it inherits the single minus sign.
4

Verify at 45 degrees

$\sin 135° = \tfrac{1}{\sqrt{2}} = \sin 45° \ \checkmark$

$\cos 135° = -\tfrac{1}{\sqrt{2}} = -\cos 45° \ \checkmark$

$\tan 135° = -1 = -\tan 45° \ \checkmark$

All three hold, using the second-quadrant values from Lesson 3.

AnswerSine is preserved; cosine and tangent change sign

Watch Me Solve It · Establishing the complementary relationships

+15 XP per step

PROBLEM

Use a right-angled triangle to prove that $\sin(90° - \theta) = \cos\theta$, and verify it using the $30$-$60$-$90$ triangle.

1

Set up the triangle

In a right-angled triangle the angles sum to $180°$ and one is $90°$, so the other two sum to $90°$. Call them $\theta$ and $90° - \theta$.
2

Identify how the sides are shared

The side opposite $\theta$ is the side adjacent to $90° - \theta$, because there are only two non-hypotenuse sides and each angle faces one of them. The hypotenuse serves both angles.
3

Write the two ratios

$\sin(90° - \theta) = \frac{\text{side opposite } (90°-\theta)}{h} = \frac{\text{side adjacent to } \theta}{h} = \cos\theta$

The middle step is the observation from the previous line: the same physical side has two descriptions.
4

Verify with the special triangle

$\sin 60° = \tfrac{\sqrt{3}}{2}, \qquad \cos 30° = \tfrac{\sqrt{3}}{2}$

And $60° = 90° - 30°$, so the relationship holds. The companion $\sin 30° = \cos 60° = \tfrac{1}{2}$ checks the other direction.

Answer$\sin(90° - \theta) = \cos\theta$, since one angle's opposite side is the other's adjacent side

Watch Me Solve It · Using the relationships

+15 XP per step

PROBLEM

(a) Express $\cos 162°$ in terms of an acute angle. (b) Simplify $\dfrac{\sin 35°}{\cos 55°}$. (c) Given $\sin\theta = 0.4$ with $0° \leq \theta \leq 180°$, find both possible values of $\theta$ to the nearest degree.

1

(a) Use the supplementary relationship

$\cos 162° = \cos(180° - 18°) = -\cos 18°$

The name stays cosine and the sign flips, since $162°$ is in the second quadrant where cosine is negative.
2

(b) Look for a complementary pair

$35° + 55° = 90°$

They are complementary, so $\cos 55° = \sin 35°$.
3

(b) Substitute and cancel

$\frac{\sin 35°}{\sin 35°} = 1$

No calculator is needed at all: spotting the pair removed the calculation.
4

(c) Find both solutions

$\theta \approx 23.6° \ \Rightarrow \ 24°$

$180° - 23.6° = 156.4° \ \Rightarrow \ 156°$

The calculator gives the acute solution. The supplement is the second, because supplementary angles share a sine, and both lie in the stated interval so both must be given.

Answer(a) $-\cos 18°$; (b) $1$; (c) $\theta \approx 24°$ or $156°$

Brain Trainer · 5 problems

Brain Trainer · Supplements and complements

5 problems

Five items on the two relationships. Work each one, then reveal the answer.

1 Simplify $\sin(180° - 40°)$.

Supplementary angles share a sine.$\sin 40°$
2 Simplify $\cos(180° - 40°)$.

The name stays; the sign flips.$-\cos 40°$
3 Simplify $\cos(90° - 25°)$.

Complementary angles swap the names.$\sin 25°$
4 $\sin\theta = \sin 70°$ and $0° \le \theta \le 180°$. Find both values of $\theta$.

The angle itself and its supplement.$70°$ and $110°$
5 Simplify $\dfrac{\cos 18°}{\sin 72°}$.

$18 + 72 = 90$, so they are complementary.$1$

Complete in your workbook.

Multiple Choice · 5 questions

MC1

Supplementary sine

+10 XP

For any angle $\theta$, $\sin(180° - \theta)$ equals:

MC2

Supplementary cosine

+10 XP

For any angle $\theta$, $\cos(180° - \theta)$ equals:

MC3

Complementary

+10 XP

$\cos 63°$ is equal to:

MC4

Two solutions

+10 XP

If $\sin\theta = 0.7$ and $0° \leq \theta \leq 180°$, the number of solutions is:

MC5

Telling them apart

+10 XP

Which statement is correct?

Short Answer · 3 questions

Establish both pairs

+15 XP

SHORT ANSWER

(a) Using the unit circle, establish the three supplementary relationships for sine, cosine and tangent.
(b) Using a right-angled triangle, establish that $\sin(90° - \theta) = \cos\theta$ and $\cos(90° - \theta) = \sin\theta$.
(c) Verify one relationship from each pair using exact values.
(d) State clearly what distinguishes the two pairs.

Write your working in your book.

Apply them

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SHORT ANSWER

(a) Express $\sin 154°$, $\cos 154°$ and $\tan 154°$ in terms of ratios of an acute angle.
(b) Simplify $\dfrac{\cos 40°}{\sin 50°} + \dfrac{\sin 130°}{\sin 50°}$ without a calculator.
(c) Given $\cos\theta = -0.35$ with $0° \leq \theta \leq 180°$, explain why there is exactly one solution, unlike the sine case.
(d) Given $\sin\theta = 0.28$ with $0° \leq \theta \leq 180°$, find both solutions to the nearest degree.

Write your working in your book.

Reasoning with the identities

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SHORT ANSWER

(a) Explain why $\sin 45°$ must equal $\cos 45°$, using a relationship rather than the special triangle.
(b) Explain why $\sin 90°$ has no distinct supplementary partner, and what happens to the second solution in that case.
(c) Show that $\sin^2 20° + \sin^2 70° = 1$, and explain which relationship you used.
(d) A student writes $\cos(90° - \theta) = -\sin\theta$. Give a counterexample and explain the error.

Write your working in your book.

Stretch Challenge · How far the relationships reach

+25 XP

CHALLENGE

(a) The complementary relationships were proved with a right-angled triangle, which requires $\theta$ to be acute. Show they still hold for $\theta = 120°$ by checking directly, and explain why the unit circle gives a proof that covers every angle.
(b) Establish a relationship for $\tan(90° - \theta)$ and state a restriction on $\theta$.
(c) Combine the supplementary and complementary relationships to express $\sin(90° + \theta)$ in terms of a ratio of $\theta$, and verify your answer at $\theta = 30°$.

(a) Check at $\theta = 120°$: the complement is $90° - 120° = -30°$. Then $\cos(-30°) = \cos 30° = \tfrac{\sqrt{3}}{2}$, using the negative-angle rule from Lesson 3, and $\sin 120° = \tfrac{\sqrt{3}}{2}$. They agree, so $\cos(90° - \theta) = \sin\theta$ holds here even though no right-angled triangle contains a $120°$ angle. The unit circle proves it in general: turning through $90° - \theta$ can be seen as reflecting the point at $\theta$ in the line at $45°$, and reflecting in that line exchanges the two coordinates. Since sine and cosine are the two coordinates, exchanging them is exactly the statement that each becomes the other. That argument never mentions a triangle, so it holds for every angle, including obtuse and negative ones.

(b) Dividing the two complementary relationships: $\tan(90° - \theta) = \dfrac{\sin(90° - \theta)}{\cos(90° - \theta)} = \dfrac{\cos\theta}{\sin\theta}$, which is the reciprocal of $\tan\theta$. So $\tan(90° - \theta) = \dfrac{1}{\tan\theta}$. The restriction is that $\tan\theta$ must be neither zero nor undefined, since the reciprocal of zero does not exist and the left side is itself undefined where $\cos(90°-\theta) = 0$. That rules out $\theta = 0°$ and $\theta = 90°$, and every angle differing from those by a multiple of $180°$. This is the reciprocal relationship seen in Lesson 2 with $\tan 30°$ and $\tan 60°$, now proved in general.

(c) Write $90° + \theta = 180° - (90° - \theta)$, which can be checked by expanding the right side: $180° - 90° + \theta = 90° + \theta$. Now apply the supplementary relationship for sine, which preserves it: $\sin(90° + \theta) = \sin(90° - \theta)$. Then apply the complementary relationship: $\sin(90° - \theta) = \cos\theta$. So $\sin(90° + \theta) = \cos\theta$. Verifying at $\theta = 30°$: $\sin 120° = \tfrac{\sqrt{3}}{2}$ and $\cos 30° = \tfrac{\sqrt{3}}{2}$, which agree. Note this is the same relationship as the graph shift in Lesson 4, where $\cos\theta = \sin(\theta + 90°)$: the identity derived here algebraically is the translation seen there geometrically.

Quick Review

recap

Supplementary

$\sin$ kept; $\cos$ and $\tan$ negated

Complementary

$\sin$ and $\cos$ exchange; no sign change

Two solutions

A given sine has an acute answer and its supplement

Test it

Check any identity at $30°$ or $60°$ before using it

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