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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
If $\cos\alpha+\cos\beta=\frac{1}{3}$ and $\sin\alpha+\sin\beta=\frac{1}{4},$ prove that $\cos\frac{\alpha-\beta}{2}=\pm\frac{5}{24}$
✓

Answer

We have,
$\cos\alpha+\cos\beta=\frac{1}{3}$ and $\sin\alpha+\sin\beta=\frac{1}{4}$
Squaring and adding, we get
$(\cos^2\alpha+\cos^2\beta+2\cos\alpha\cos\beta)+(\sin^2\alpha+\sin^2\beta+2\sin\alpha\sin\beta)=\frac{1}{9}+\frac{1}{16}$
$\Rightarrow1+1+2(\cos\alpha\cos\beta+\sin\alpha\sin\beta)=\frac{25}{144}$
$\Rightarrow2\cos(\alpha-\beta)=\frac{25}{144}-2=\frac{-263}{144}$
$\Rightarrow\cos(\alpha-\beta)=\frac{-263}{288}$
Now,
$\cos\Big(\frac{\alpha-\beta}{2}\Big)=\sqrt{\frac{1+\cos(\alpha-\beta)}{2}}$
$=\sqrt{\frac{1-\frac{263}{288}}{2}}=\sqrt{\frac{25}{576}}$
$=\pm\frac{5}{24}$
$\therefore\cos\Big(\frac{\alpha-\beta}{2}\Big)=\pm\frac{5}{25}$

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