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Toolkit 37

Trigonometric Transformations

sin⁡(90∘−α)=cos⁡αsin⁡(90∘+α)=cos⁡α\sin(90^\circ-\alpha)=\cos\alpha\qquad \sin(90^\circ+\alpha)=\cos\alpha
cos⁡(90∘−α)=sin⁡αcos⁡(90∘+α)=−sin⁡α\cos(90^\circ-\alpha)=\sin\alpha\qquad \cos(90^\circ+\alpha)=-\sin\alpha
tan⁡(90∘−α)=cot⁡αtan⁡(90∘+α)=−cot⁡α\tan(90^\circ-\alpha)=\cot\alpha\qquad \tan(90^\circ+\alpha)=-\cot\alpha
cot⁡(90∘−α)=tan⁡αcot⁡(90∘+α)=−tan⁡α\cot(90^\circ-\alpha)=\tan\alpha\qquad \cot(90^\circ+\alpha)=-\tan\alpha
sin⁡(180∘−α)=sin⁡αsin⁡(180∘+α)=−sin⁡α\sin(180^\circ-\alpha)=\sin\alpha\qquad \sin(180^\circ+\alpha)=-\sin\alpha
cos⁡(180∘−α)=−cos⁡αcos⁡(180∘+α)=−cos⁡α\cos(180^\circ-\alpha)=-\cos\alpha\qquad \cos(180^\circ+\alpha)=-\cos\alpha
tan⁡(180∘−α)=−tan⁡αtan⁡(180∘+α)=tan⁡α\tan(180^\circ-\alpha)=-\tan\alpha\qquad \tan(180^\circ+\alpha)=\tan\alpha
cot⁡(180∘−α)=−cot⁡αcot⁡(180∘+α)=cot⁡α\cot(180^\circ-\alpha)=-\cot\alpha\qquad \cot(180^\circ+\alpha)=\cot\alpha
sin⁡(−α)=−sin⁡α\sin(-\alpha)=-\sin\alpha
cos⁡(−α)=cos⁡α\cos(-\alpha)=\cos\alpha
tan⁡(−α)=−tan⁡α\tan(-\alpha)=-\tan\alpha
cot⁡(−α)=−cot⁡α\cot(-\alpha)=-\cot\alpha

Proof

Unit circle showing 180°−α, 180°+α, and −α reflections

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

sin⁡(180∘−α)=sin⁡α,cos⁡(180∘−α)=−cos⁡α,tan⁡(180∘−α)=−tan⁡α,cot⁡(180∘−α)=−cot⁡α,\begin{aligned} \sin(180^\circ-\alpha) &= \sin\alpha, & \cos(180^\circ-\alpha) &= -\cos\alpha, \\ \tan(180^\circ-\alpha) &= -\tan\alpha, & \cot(180^\circ-\alpha) &= -\cot\alpha, \end{aligned}
sin⁡(180∘+α)=−sin⁡α,cos⁡(180∘+α)=−cos⁡α,tan⁡(180∘+α)=tan⁡α,cot⁡(180∘+α)=cot⁡α,\begin{aligned} \sin(180^\circ+\alpha) &= -\sin\alpha, & \cos(180^\circ+\alpha) &= -\cos\alpha, \\ \tan(180^\circ+\alpha) &= \tan\alpha, & \cot(180^\circ+\alpha) &= \cot\alpha, \end{aligned}
sin⁡(−α)=−sin⁡α,cos⁡(−α)=cos⁡α,tan⁡(−α)=−tan⁡α,cot⁡(−α)=−cot⁡α.\begin{aligned} \sin(-\alpha) &= -\sin\alpha, & \cos(-\alpha) &= \cos\alpha, \\ \tan(-\alpha) &= -\tan\alpha, & \cot(-\alpha) &= -\cot\alpha. \end{aligned}

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

Unit circle with congruent right triangles OAB and OCD illustrating 90°−α

Let A=(cos⁡α,sin⁡α)A=(\cos\alpha,\sin\alpha) be the point on the unit circle corresponding to α\alpha, and let CC be the point corresponding to 90∘−α90^\circ-\alpha. Drop perpendiculars from AA and CC to the xx- and yy-axes at BB and DD respectively. The right triangles OABOAB and OCDOCD share the hypotenuse OA=OC=1OA=OC=1 and have equal acute angles α\alpha at OO, so they are congruent. Hence OD=OB=cos⁡αOD = OB = \cos\alpha and DC=AB=sin⁡αDC = AB = \sin\alpha, which are the coordinates of CC:

(cos⁡(90∘−α),sin⁡(90∘−α))=(sin⁡α,cos⁡α).(\cos(90^\circ-\alpha), \sin(90^\circ-\alpha)) = (\sin\alpha, \cos\alpha).

Therefore,

sin⁡(90∘−α)=cos⁡α,cos⁡(90∘−α)=sin⁡α.\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∘−α)=sin⁡(90∘−α)cos⁡(90∘−α)=cos⁡αsin⁡α=cot⁡α,\tan(90^\circ-\alpha) = \frac{\sin(90^\circ-\alpha)}{\cos(90^\circ-\alpha)} = \frac{\cos\alpha}{\sin\alpha} = \cot\alpha,
cot⁡(90∘−α)=cos⁡(90∘−α)sin⁡(90∘−α)=sin⁡αcos⁡α=tan⁡α.\cot(90^\circ-\alpha) = \frac{\cos(90^\circ-\alpha)}{\sin(90^\circ-\alpha)} = \frac{\sin\alpha}{\cos\alpha} = \tan\alpha.

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

sin⁡(90∘+α)=sin⁡(180∘−(90∘−α))=sin⁡(90∘−α)=cos⁡α,cos⁡(90∘+α)=cos⁡(180∘−(90∘−α))=−cos⁡(90∘−α)=−sin⁡α.\begin{aligned} \sin(90^\circ+\alpha) &= \sin\bigl(180^\circ-(90^\circ-\alpha)\bigr) = \sin(90^\circ-\alpha) = \cos\alpha, \\ \cos(90^\circ+\alpha) &= \cos\bigl(180^\circ-(90^\circ-\alpha)\bigr) = -\cos(90^\circ-\alpha) = -\sin\alpha. \end{aligned}

Dividing then yields

tan⁡(90∘+α)=−cot⁡α,cot⁡(90∘+α)=−tan⁡α.□\begin{aligned} \tan(90^\circ+\alpha) &= -\cot\alpha, \\ \cot(90^\circ+\alpha) &= -\tan\alpha. \quad\square \end{aligned}