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

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Ratios Of Compound Angles3 Marks
Question
Prove that:
$\big(\cos\alpha+\cos\beta^2\big)+\big(\sin\alpha+\sin\beta\big)^2=2\cos^2\Big(\frac{\alpha-\beta}{2}\Big)$
✓

Answer

$\text{LHS}=\big(\cos\lambda+\cos\beta\big)^2+\big(\sin\lambda+\sin\beta\big)^2$
$=\cos^2\lambda+\cos^2\beta+2\cos\lambda\cos\beta+\sin^2\lambda+\sin^2\beta+2\sin\lambda+\sin\beta$
$=\Big(\cos^2\lambda+\sin^2\lambda\Big)+\Big(\cos^2\beta+\sin^2\beta\Big)+2\Big(\cos\lambda\cos\beta+\sin\lambda\sin\beta\Big)$
$=1+1+2\cos(\lambda-\beta)$
$=2+2\cos(\lambda-\beta)$
$=2\big(1+\cos(\lambda-\beta)\big)$
$=2.2\cos^2\Big(\frac{\lambda-\beta}{2}\Big)$
$=4\cos^2\Big(\frac{\lambda-\beta}{2}\Big)=\text{RHS}$

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