Rule

Cofunction Identities

For any angle θ, the following trigonometric identities hold true.

sin ( π/2 - θ ) = cosθ

\cos\left(\dfrac \pi 2 - \theta\right) = \sin\theta

\tan\left(\dfrac \pi 2 - \theta\right) = \cot\theta

Proof

For Acute Angles
Consider a right triangle. The measure of its right angle is 90^(∘) or π2 radians. Let θ be the radian measure of one of the acute angles. Since the sum of two acute angles in a right triangle is π2, the measure of the third acute angle must be π2-θ.

Let also a, b, and c represent the side lengths of the triangle. In this case, cosine of θ can be expressed as the ratio of the lengths of the angle's adjacent side and the hypotenuse. cosθ = b/c At the same time, sine of the opposite angle can be expressed as the ratio of the lengths of the angle's opposite side and the hypotenuse. sin ( π/2 - θ ) = b/c Since the right-hand sides of the equations are equal, by the Transitive Property of Equality, the left-hand sides are also equal.

sin ( π/2 - θ ) = cosθ

This identity is true for all angles, not just those that make it possible to construct a right triangle. Using similar reasoning, the corresponding identities for cosine and tangent can be proven.

Proof

For Any Angle
The given identities can be proven using the Angle Sum and Difference Identities for any angle measure. Consider the identity for the sine of the difference between two angles. sin(x-y)=sinxcosy-cosxsiny This identity will be applied to the left-hand side of the first identity.

sin ( π/2 - θ )

sin(α-β)=sin(α)cos(β)-cos(α)sin(β)

sin ( π/2)cos(θ)-cos ( π/2)sin(θ)

\ifnumequal{90}{0}{\sin\left(0\right)=0}{}\ifnumequal{90}{30}{\sin\left(\dfrac{\pi}{6}\right)=\dfrac{1}{2}}{}\ifnumequal{90}{45}{\sin\left(\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{90}{60}{\sin\left(\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{90}{90}{\sin\left(\dfrac{\pi}{2}\right)=1}{}\ifnumequal{90}{120}{\sin\left(\dfrac{2\pi}{3}\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{90}{135}{\sin\left(\dfrac{3\pi}{4}\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{90}{150}{\sin\left(\dfrac{5\pi}{6}\right)=\dfrac{1}{2}}{}\ifnumequal{90}{180}{\sin\left(\pi\right)=0}{}\ifnumequal{90}{210}{\sin\left(\dfrac{7\pi}6\right)=\text{-} \dfrac 1 2}{}\ifnumequal{90}{225}{\sin\left(\dfrac{5\pi}{4}\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{90}{240}{\sin\left(\dfrac{4\pi}3\right)=\text{-} \dfrac {\sqrt 3}2}{}\ifnumequal{90}{270}{\sin\left(\dfrac{3\pi}{2}\right)=\text{-} 1}{}\ifnumequal{90}{300}{\sin\left(\dfrac{5\pi}3\right)=\text{-} \dfrac {\sqrt 3}2}{}\ifnumequal{90}{315}{\sin\left(\dfrac{7\pi}4\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{90}{330}{\sin\left(\dfrac{11\pi}6\right)=\text{-} \dfrac 1 2}{}\ifnumequal{90}{360}{\sin\left(2\pi\right)=0}{}

1*cos(θ)-cos ( π/2)sin(θ)
cos(θ)-cos ( π/2)sin(θ)

\ifnumequal{90}{0}{\cos\left(0\right)=1}{}\ifnumequal{90}{30}{\cos\left(\dfrac{\pi}{6}\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{90}{45}{\cos\left(\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{90}{60}{\cos\left(\dfrac{\pi}{3}\right)=\dfrac{1}{2}}{}\ifnumequal{90}{90}{\cos\left(\dfrac{\pi}{2}\right)=0}{}\ifnumequal{90}{120}{\cos\left(\dfrac{2\pi}{3}\right)=\text{-} \dfrac{1}{2}}{}\ifnumequal{90}{135}{\cos\left(\dfrac{3\pi}{4}\right)=\text{-} \dfrac{\sqrt{2}}{2}}{}\ifnumequal{90}{150}{\cos\left(\dfrac{5\pi}{6}\right)=\text{-} \dfrac{\sqrt{3}}{2}}{}\ifnumequal{90}{180}{\cos\left(\pi\right)=\text{-} 1}{}\ifnumequal{90}{210}{\cos\left(\dfrac{7\pi}6\right)=\text{-} \dfrac{\sqrt 3}2}{}\ifnumequal{90}{225}{\cos\left(\dfrac{5\pi}{4}\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{90}{240}{\cos\left(\dfrac{4\pi}3\right)=\text{-} \dfrac {1}2}{}\ifnumequal{90}{270}{\cos\left(\dfrac{3\pi}{2}\right)=0}{}\ifnumequal{90}{300}{\cos\left(\dfrac{5\pi}3\right)=\dfrac{1}2}{}\ifnumequal{90}{315}{\cos\left(\dfrac{7\pi}4\right)=\dfrac {\sqrt{2}} {2}}{}\ifnumequal{90}{330}{\cos\left(\dfrac{11\pi}6\right)=\dfrac{\sqrt 3}2}{}\ifnumequal{90}{360}{\cos\left(2\pi\right)=1}{}

cos(θ)-0*sin(θ)
cos(θ)-0
cos(θ)

Therefore, the Cofunction Identity for sine was obtained.

sin ( π/2 - θ )=cos(θ)

Exercises
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