Question

Prove More Reduction Formulas, By the Angle-Side-Angle Theorem from elementary geometry, triangles $CDO$ and $AOB$in the figure to the right are congruent. Explain how this proves that if $B$has coordinates $\left(x,y\right)$, then $D$has coordinates $\left(-y,x\right)$. Then explain how the figure shows that the following reduction formulas are valid.

$\sin \left(t+\frac{\mathrm{\pi }}{2}\right)=\cos t$

$\cos \left(t+\frac{\mathrm{\pi }}{2}\right)=-\sin t$

$\tan \left(t+\frac{\mathrm{\pi }}{2}\right)=-\cot t$

Short answer

It is proved that

$\mathbf{sin}\mathbf{(}\mathbf{t}\mathbf{+}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{)}\mathbf{=}\mathbf{cos}\mathit{t}$

$\mathbf{cos}\left(t+\frac{\pi }{2}\right)\mathbf{=}\mathbf{-}\mathbf{sin}\mathit{t}$

$\mathbf{tan}\left(t+\frac{\pi }{2}\right)\mathbf{=}\mathbf{-}\mathbf{cot}\mathit{t}$

Step-by-step solution

Step 1. Given Information

Step 2. Proof.

Now, apply the sine formula over the right side $\sin \left(t+\frac{\pi }{2}\right)$.

$\sin \left(t+\frac{\pi }{2}\right)=\sin t\cos \frac{\pi }{2}+\cos t\sin \frac{\pi }{2}$

$$\begin{gathered} =0+\cos t \\ =\cos t \end{gathered}$$

Now, apply the cosine formula over the right side $\cos \left(t+\frac{\pi }{2}\right)$.

$\cos \left(t+\frac{\pi }{2}\right)=\cos t\cos \frac{\pi }{2}-\sin t\sin \frac{\pi }{2}$

$$\begin{gathered} =0-\sin t \\ =-\sin t \end{gathered}$$

Now, express $\tan \left(t+\frac{\pi }{2}\right)$in terms of sine cosine and solve

$tan\left(t+\frac{\pi }{2}\right)=\frac{sin\left(t+\frac{\pi }{2}\right)}{cos\left(t+\frac{\pi }{2}\right)}$

$$\begin{gathered} =\frac{cost}{-sint} \\ =-cott \end{gathered}$$