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Trigonometric Ratios Of Multiple And Sub Multiple Angles — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Ratios Of Multiple And Sub Multiple Angles4 Marks
Question
If $\sin\alpha+\sin\beta=\text{a}$ and $\cos\alpha+\cos\beta=\text{b},$ show that
$\cos(\alpha+\beta)=\frac{\text{b}^2-\text{a}^2}{\text{b}^2+\text{a}^2}$
✓

Answer

$\text{a}^2+\text{b}^2=(\sin\alpha+\sin\beta)^2+(\cos\alpha+\cos\beta)^2$
$=\sin^2\alpha+\sin^2\beta+\cos^2\alpha+\cos^2\beta+2\sin\alpha\sin\beta+2\cos\alpha\cos\beta$
$=2+2\cos(\alpha+\beta)$
$\Rightarrow\text{b}^2+\text{a}^2=(\cos\alpha+\cos\beta)^2-(\sin\alpha+\sin\beta)^2$
$\Rightarrow\text{b}^2+\text{a}^2=\cos^2\alpha+\cos^2\beta-\sin^2\alpha+\sin^2\beta+2\cos\alpha\cos\beta-2\sin\alpha\sin\beta$
$\Rightarrow\text{b}^2+\text{a}^2=\Big(\cos^2\alpha+\sin^2\beta\Big)+\Big(\cos^2\beta-\sin^2\alpha\Big)-2\cos(\alpha+\beta)$
$\Rightarrow\text{b}^2+\text{a}^2=2\cos(\alpha+\beta)+\cos(\alpha-\beta)+2\cos(\alpha-\beta)$
$\Rightarrow\text{b}^2+\text{a}^2=\cos(\alpha+\beta)(2+2\cos(\alpha-\beta) )$
$\Rightarrow\text{b}^2+\text{a}^2=\cos(\alpha+\beta)(\text{a}^2+\text{b}^2 )$
$\frac{\text{b}^2-\text{a}^2}{\text{b}^2+\text{a}^2}=\cos(\alpha+\beta)$

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