Question

Prove that (sin 32° cos 58° + cos 32° sin 58°) = 1.

Answer 1

(sin 32° cos 58° + cos 32° sin 58°) = 1

$$\begin{gathered} {\text{LHS=sin32}}^0\cos {58}^0+\cos {32}^0{\text{sin58}}^0 \\ \text{=sin}({90}^0-{58}^0)\cos {58}^0+\cos ({90}^0-{58}^0){\text{sin58}}^0 \\ \text{=}\cos {58}^0\times \cos {58}^0+{\text{sin58}}^0\times {\text{sin58}}^0\left[\begin{array}{l}\because \text{sin}({90}^0-\theta )=\cos \theta ,\\ \cos ({90}^0-\theta )=\cos \theta \end{array}\right] \\ \text{=}{\cos }^2{58}^0+{\text{sin}}^2{\text{58}}^0 \\ \text{=1}[\because {\text{sin}}^2\theta +{\cos }^2\theta =1] \\ \text{=RHS} \end{gathered}$$