# 10.02 Cofunction identities

Lesson

Recall that in a right triangle, the two acute angles together make a right-angle. We say the acute angles are complementary (to one another). If the two acute angles have measures $\alpha$α and $\beta$β, then $\alpha+\beta=90^\circ$α+β=90° and so, $\beta=90^\circ-\alpha$β=90°α.

The diagram below illustrates the following relationships.

$\cos\alpha=\frac{b}{h}=\sin\beta=\sin\left(90^\circ-\alpha\right)$cosα=bh=sinβ=sin(90°α)

$\sin\alpha=\frac{a}{h}=\cos\beta=\cos\left(90^\circ-\alpha\right)$sinα=ah=cosβ=cos(90°α)

$\cot\alpha=\frac{b}{a}=\tan\beta=\tan\left(90^\circ-\alpha\right)$cotα=ba=tanβ=tan(90°α)

Thus, the 'co' in complementary explains the meaning of cosine in relation to sine, and to cotangent in relation to tangent.

The relationships above are identities because they are true whatever the value of the angle $\alpha$α. We can extend the relationships to reciprocal trig functions to obtain the following set of cofunction identities:

Cofunction identities
 $\sin\alpha\equiv\cos\left(90^\circ-\alpha\right)$sinα≡cos(90°−α) $\tan\alpha\equiv\cot\left(90^\circ-\alpha\right)$tanα≡cot(90°−α) $\sec\alpha\equiv\csc\left(90^\circ-\alpha\right)$secα≡csc(90°−α) $\cos\alpha\equiv\sin\left(90^\circ-\alpha\right)$cosα≡sin(90°−α) $\cot\alpha\equiv\tan\left(90^\circ-\alpha\right)$cotα≡tan(90°−α) $\csc\alpha\equiv\sec\left(90^\circ-\alpha\right)$cscα≡sec(90°−α)

These identities are true not only in right triangle trigonometry, but they also hold for angles of any size. This can be confirmed by thinking about the geometry in the unit circle diagram that is used for defining the trigonometric functions of angles of any magnitude.

#### Worked example

##### Question 1

Simplify the relation $\sin\left(90^\circ-\theta\right)=\sqrt{3}\sin\theta$sin(90°θ)=3sinθ where $0^\circ<\theta<360^\circ$0°<θ<360° is an acute angle.

Think: Rearrange the equation so that the trigonometric functions are on one side and the coefficients are on the other.

Do:

 $\sin\left(90^\circ-\theta\right)$sin(90°−θ) $=$= $\sqrt{3}\sin\theta$√3sinθ Given statement $\cos\theta$cosθ $=$= $\sqrt{3}\sin\theta$√3sinθ Using the cofunction relationships $1$1 $=$= $\frac{\sqrt{3}\sin\theta}{\cos\theta}$√3sinθcosθ​ Dividing both sides by $\cos\theta$cosθ $1$1 $=$= $\sqrt{3}\tan\theta$√3tanθ Using the fact that $\tan\theta=\frac{\sin\theta}{\cos\theta}$tanθ=sinθcosθ​ $\tan\theta$tanθ $=$= $\frac{1}{\sqrt{3}}$1√3​ Dividing both sides by $\sqrt{3}$√3

We recognize an exact value for $\tan$tan and conclude that $\theta=30^\circ$θ=30° if $\theta$θ is acute.

Reflect: We should check that there is also a third quadrant solution, $\theta=210^\circ$θ=210°.

#### Practice questions

##### Question 2

Simplify the following expression using complementary angles:

$\frac{\sin51^\circ}{\cos39^\circ}$sin51°cos39°

##### Question 3

Rewrite $\sec\frac{\pi}{9}$secπ9 in terms of its cofunction.

##### Question 4

Simplify $\sin\left(90^\circ-y\right)\times\tan y$sin(90°y)×tany.