Question
Prove that $(\sin32^\circ\cos58^\circ+\cos32^\circ\sin58^\circ)=1.$
✓
Answer
$(\sin32^\circ\cos58^\circ+\cos32^\circ\sin58^\circ)=1$$\text{LHS}=\sin32^\circ\cos58^\circ+\cos32^\circ\sin58^\circ$
$=\sin(90^\circ-58^\circ)\cos58^\circ+\cos(90^\circ-58^\circ)\sin58^\circ$
$=\cos58^\circ\times\cos58^\circ+\sin58^\circ\times\sin58^\circ$ $\begin{bmatrix}\because\sin(90^\circ-\theta)=\cos\theta,\\\cos(90^\circ-\theta)=\cos\theta\end{bmatrix}$
$=\cos^258^\circ+\sin^258^\circ$
$=1$ $\big[\because\sin^2\theta+\cos^2\theta=1\big]$
$=\text{RHS}$
$=\sin(90^\circ-58^\circ)\cos58^\circ+\cos(90^\circ-58^\circ)\sin58^\circ$
$=\cos58^\circ\times\cos58^\circ+\sin58^\circ\times\sin58^\circ$ $\begin{bmatrix}\because\sin(90^\circ-\theta)=\cos\theta,\\\cos(90^\circ-\theta)=\cos\theta\end{bmatrix}$
$=\cos^258^\circ+\sin^258^\circ$
$=1$ $\big[\because\sin^2\theta+\cos^2\theta=1\big]$
$=\text{RHS}$
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