Question

The value of $\sin 16°+\cos 16°$ is

• A. $\frac{1}{\sqrt{3}}\left(\sqrt{2}\cos 1°+\sin 1°\right)$
• B. $\frac{1}{\sqrt{2}}\left(\cos 1°+\sqrt{3}\sin 1°\right)$
• C. $\frac{1}{\sqrt{3}}\left(\cos 1°+\sqrt{2}\sin 1°\right)$
• D. $\frac{1}{\sqrt{2}}\left(\sqrt{3}\cos 1°+\sin 1°\right)$

Answer 1

The correct option is D

$\frac{1}{\sqrt{2}}\left(\sqrt{3}\cos 1°+\sin 1°\right)$

Explanation for the correct option:

Find the value of the expression:

$$\begin{gathered} \sin \left(A+B\right)=\sin A\cos B+\cos B\sin A \\ \cos \left(A+B\right)=\cos A\cos B-\sin A\sin B \end{gathered}$$

Use the above formula in the $\sin 16°+\cos 16°$

$\begin{array}{rcl}\sin 16°+\cos 16°& =& \sin \left(15°+1°\right)+\cos \left(15°+1°\right)\\ & =& \left(\sin 15°\cos 1°+\cos 15°\sin 1°\right)+\left(\cos 15°\cos 1°-\sin 15°\sin 1°\right)\\ & =& \sin 1°\left(\cos 15°-\sin 15°\right)+\cos 1°\left(\sin 15°+\cos 15°\right)...\left(1\right)\end{array}$

Now,

$\begin{array}{rcl}\cos 15°& =& \cos \left(45°-30°\right)\\ & =& \cos 45°\cos 30°+\sin 45°\sin 30°\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{cos}\left(A-B\right)\mathbf{=}\mathbf{cos}\mathbf{A}\mathbf{cos}\mathbf{B}\mathbf{-}\mathbf{sin}\mathbf{A}\mathbf{sin}\mathbf{B}\mathbf{}\mathbf{]}\\ & =& \frac{1}{\sqrt{2}}·\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}·\frac{1}{2}\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{sin}\mathbf{45}\mathbf{°}\mathbf{=}\mathbf{cos}\mathbf{45}\mathbf{°}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\mathbf{,}\mathbf{}\mathbf{sin}\mathbf{30}\mathbf{°}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{,}\mathbf{}\mathbf{cos}\mathbf{30}\mathbf{°}\mathbf{=}\frac{\sqrt{\mathbf{3}}}{\mathbf{2}}\mathbf{]}\\ & =& \frac{\sqrt{3}+1}{2\sqrt{2}}\end{array}$

And

$\begin{array}{rcl}\sin 15°& =& \sin \left(45°-30°\right)\\ & =& \sin 45°\cos 30°-\cos 45°\sin 30°\left[\because \sin \left(A-B\right)=\sin A\cos B-\cos A\sin B\right]\\ & =& \frac{1}{\sqrt{2}}·\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}·\frac{1}{2}\left[\because \sin 45°=\cos 45°=\frac{1}{\sqrt{2}},\sin 30°=\frac{1}{2},\cos 30°=\frac{\sqrt{3}}{2}\right]\\ & =& \frac{\sqrt{3}-1}{2\sqrt{2}}\end{array}$

Substitute the value of $\cos 15°$ and $\sin 15°$ in the equation $\left(1\right)$, we get

$\begin{array}{rcl}& \Rightarrow & \sin 16°+\cos 16°=\sin 1°\left(\frac{\sqrt{3}+1}{2\sqrt{2}}-\frac{\sqrt{3}-1}{2\sqrt{2}}\right)+\cos 1°\left(\frac{\sqrt{3}-1}{2\sqrt{2}}+\frac{\sqrt{3}+1}{2\sqrt{2}}\right)\\ & =& \sin 1°\left(\frac{\sqrt{3}+1-\sqrt{3}+1}{2\sqrt{2}}\right)+\cos 1°\left(\frac{\sqrt{3}-1+\sqrt{3}+1}{2\sqrt{2}}\right)\\ & =& \sin 1°\left(\frac{2}{2\sqrt{2}}\right)+\cos 1°\left(\frac{2\sqrt{3}}{2\sqrt{2}}\right)\\ & =& \frac{1}{\sqrt{2}}\sin 1°+\frac{\sqrt{3}}{\sqrt{2}}\cos 1°\\ & =& \frac{1}{\sqrt{2}}\left(\sin 1°+\sqrt{3}\cos 1°\right)\end{array}$

Hence, the correct option is D.