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Trigonometric Ratios Of Compound Angles — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Ratios Of Compound Angles4 Marks
Question
If $\sin\alpha+\sin\beta=\text{a}$ and $\cos\alpha+\cos\beta=\text{b}$ prove that
$\sin(\alpha+\beta)=\frac{2\text{ab}}{\text{a}^2+\text{b}^2}$
✓

Answer

we have,
$\sin\alpha+\sin\beta=\text{a}\ \&\ \cos\alpha+\cos\beta=\text{b}\ .....(\text{A})$
squaring and adding, we get
$\sin^2\alpha+\sin^2\beta+2\sin\alpha\sin\beta+\cos^2\alpha+\cos^2\beta+2\cos\alpha\cos\beta=\text{a}^2+\text{b}^2$
$\Rightarrow1+1+2(\sin\alpha\sin\beta+\cos\alpha\cos\beta)=\text{a}^2+\text{b}^2$
$\Rightarrow2(\sin\alpha\sin\beta+\cos\alpha\cos\beta)=\text{a}^2+\text{b}^2-2$
$\therefore2\cos(\alpha+\beta)=\text{a}^2+\text{b}^2-2$
Thus,
$\cos(\alpha-\beta)=\frac{\text{a}^2+\text{b}^2-2}{2}$
Again,
$\sin\alpha+\sin\beta=\text{a}\Rightarrow2\sin\frac{\alpha+\beta}{2}.\cos\frac{\alpha-\beta}{2}=\text{a}$
$\cos\alpha+\cos\beta=\text{b}\Rightarrow2\cos\frac{\alpha+\beta}{2}.\cos\frac{\alpha-\beta}{2}=\text{b}$
$\Rightarrow\tan\frac{\alpha+\beta}{2}=\frac{\text{a}}{\text{b}}\ .....\text{(B)}$
Now,
$\sin(\alpha+\beta)=\frac{1\tan\frac{\alpha+\beta}{2}}{1}+\tan^2\Big(\frac{\alpha+\beta}{2}\Big)$
thus,
$\sin(\alpha+\beta)=\frac{2\text{ab}}{\text{a}^2+\text{b}^2}$

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