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Question Number 176581 by cortano1 last updated on 22/Sep/22

$$\mathrm{Given}\{\begin{array}{l}\sin \mathrm{a}+\sin \mathrm{b}=\frac{\sqrt{2}}{2}\\ \cos \mathrm{a}+\cos \mathrm{b}=\frac{\sqrt{6}}{2}\end{array}$$$$\mathrm{for}\mathrm{a},\mathrm{b}\mathrm{real}\mathrm{numbers}.\mathrm{Evaluate}$$$$\sin (\mathrm{a}+\mathrm{b}).$$$$\left(\mathrm{A}\right)\frac{\sqrt{3}}{2}\left(\mathrm{D}\right)-\frac{\sqrt{3}}{2}$$$$\left(\mathrm{B}\right)\frac{2}{\sqrt{3}}\left(\mathrm{E}\right)-\frac{2}{\sqrt{3}}$$$$\left(\mathrm{C}\right)\frac{\sqrt{3}}{4}$$

Answered by som(math1967) last updated on 22/Sep/22

$$\frac{\sin a+sinb}{cosa+cosb}=\frac{\sqrt{2}}{\sqrt{6}}$$$$\Rightarrow tan\left(\frac{a+b}{2}\right)=\frac{1}{\sqrt{3}}$$$$sin(a+b)=\frac{2.\frac{1}{\sqrt{3}}}{1+{\left(\frac{1}{\sqrt{3}}\right)}^2}$$$$=\frac{2}{\sqrt{3}}\times \frac{3}{4}=\frac{\sqrt{3}}{2}......\left(A\right)$$

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