Trigonometric Transformations

Lesson · Intermediate

Trigonometry

\begin{aligned} \sin(90^\circ-\alpha)&=\cos\alpha &\qquad \sin(90^\circ+\alpha)&=\cos\alpha\\[10pt] \cos(90^\circ-\alpha)&=\sin\alpha & \cos(90^\circ+\alpha)&=-\sin\alpha\\[10pt] \tan(90^\circ-\alpha)&=\cot\alpha & \tan(90^\circ+\alpha)&=-\cot\alpha\\[10pt] \cot(90^\circ-\alpha)&=\tan\alpha & \cot(90^\circ+\alpha)&=-\tan\alpha\\[18pt] \sin(180^\circ-\alpha)&=\sin\alpha & \sin(180^\circ+\alpha)&=-\sin\alpha\\[10pt] \cos(180^\circ-\alpha)&=-\cos\alpha & \cos(180^\circ+\alpha)&=-\cos\alpha\\[10pt] \tan(180^\circ-\alpha)&=-\tan\alpha & \tan(180^\circ+\alpha)&=\tan\alpha\\[10pt] \cot(180^\circ-\alpha)&=-\cot\alpha & \cot(180^\circ+\alpha)&=\cot\alpha\\[18pt] \sin(-\alpha)&=-\sin\alpha\\[10pt] \cos(-\alpha)&=\cos\alpha\\[10pt] \tan(-\alpha)&=-\tan\alpha\\[10pt] \cot(-\alpha)&=-\cot\alpha \end{aligned}

Unit circle diagram for trigonometric transformations

From the unit-circle diagram above, we immediately obtain, whenever the expressions are defined,

\begin{aligned} \sin(180^\circ-\alpha)&=\sin\alpha &\qquad \cos(180^\circ-\alpha)&=-\cos\alpha\\ \tan(180^\circ-\alpha)&=-\tan\alpha & \cot(180^\circ-\alpha)&=-\cot\alpha\\[10pt] \sin(180^\circ+\alpha)&=-\sin\alpha & \cos(180^\circ+\alpha)&=-\cos\alpha\\ \tan(180^\circ+\alpha)&=\tan\alpha & \cot(180^\circ+\alpha)&=\cot\alpha\\[10pt] \sin(-\alpha)&=-\sin\alpha & \cos(-\alpha)&=\cos\alpha\\ \tan(-\alpha)&=-\tan\alpha & \cot(-\alpha)&=-\cot\alpha \end{aligned}

For the transformations involving $90^\circ$ , we use the following figure.

Unit circle diagram for 90 degree trigonometric transformations

Let $A=(\cos\alpha,\sin\alpha)$ be the point on the unit circle corresponding to $\alpha$ , and let $C$ be the point corresponding to $90^\circ-\alpha$ . Drop perpendiculars from $A$ and $C$ to the $x$ - and $y$ -axes at $B$ and $D$ respectively. The right triangles $OAB$ and $OCD$ share the hypotenuse $OA=OC=1$ and have equal acute angles $\alpha$ at $O$ , so they are congruent. Hence $OD=OB=\cos\alpha$ and $DC=AB=\sin\alpha$ , which are the coordinates of $C$ :

(\cos(90^\circ-\alpha),\sin(90^\circ-\alpha)) = (\sin\alpha,\cos\alpha)

Therefore,

\sin(90^\circ-\alpha)=\cos\alpha \qquad \cos(90^\circ-\alpha)=\sin\alpha

Using the definitions of tangent and cotangent, whenever the expressions are defined,

\tan(90^\circ-\alpha) = \frac{\sin(90^\circ-\alpha)}{\cos(90^\circ-\alpha)} = \frac{\cos\alpha}{\sin\alpha} = \cot\alpha

\cot(90^\circ-\alpha) = \frac{\cos(90^\circ-\alpha)}{\sin(90^\circ-\alpha)} = \frac{\sin\alpha}{\cos\alpha} = \tan\alpha

Finally, writing $90^\circ+\alpha=180^\circ-(90^\circ-\alpha)$ and combining the identities already proved gives, whenever the expressions are defined,

\sin(90^\circ+\alpha) = \sin(180^\circ-(90^\circ-\alpha)) = \sin(90^\circ-\alpha) = \cos\alpha

\cos(90^\circ+\alpha) = \cos(180^\circ-(90^\circ-\alpha)) = -\cos(90^\circ-\alpha) = -\sin\alpha

Dividing then yields

\tan(90^\circ+\alpha)=-\cot\alpha

\cot(90^\circ+\alpha)=-\tan\alpha