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Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions2 Marks
Question
Prove that: cot 4x (sin 5x + sin 3x) = cotx(sin 5x - sin 3x)
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Answer

We have L.H.S. = cot 4x (sin 5x + sin 3x)
$ = \frac{{\cos 4x}}{{\sin 4x}}\left[ {2\sin \left( {\frac{{5x + 3x}}{2}} \right)\cos \left( {\frac{{5x - 3x}}{2}} \right)} \right]$
$ = \frac{{\cos 4x}}{{\sin 4x}}[2\sin 4x\cos x]$= 2 cos 4x cos x
We have R.H.S. = cot x[sin5x - sin3x]
$ = \frac{{\cos x}}{{\sin x}}\left[ {2\cos \left( {\frac{{5x + 3x}}{2}} \right)\sin \left( {\frac{{5x - 3x}}{2}} \right)} \right]$
$\left[ {\because \sin C - \sin D = 2\cos \frac{{C + D}}{2} \cdot \sin \frac{{C - D}}{2}} \right]$
$ = \frac{{\cos x}}{{\sin x}}$ [2 cos 4x sin x] = 2 cos 4x cos x
Hence L.H.S. = R.H.S

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