If the expression $4 s i n 5 α c o s 3 α c o s 2 α$ is expressed as the sum of three sines then two of them are $s i n 4 α$ and $s i n 10 α$ . The third one is

s i n 8 α

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s i n 6 α

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s i n 5 α

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s i n 12 α

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Solution

The correct option is A $s i n 6 α$
Let $T = 4 sin 5 α cos 3 α cos 2 α$
Using $2 sin A cos B = sin (A + B) + sin (A - B)$
We have $2 sin 5 α cos 2 α = sin 7 α + sin 3 α$
$∴ T = 2 cos 3 α (sin 7 α + sin 3 α)$
$= 2 sin 7 α cos α + 2 sin 3 α cos 3 α$
$= sin 10 α + sin 4 α + sin 6 α (∵ 2 sin A cos A = sin 2 A)$

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Similar questions

The value of $\frac{s i n 5 α - s i n 3 α}{c o s 5 α + 2 c o s 4 α + c o s 3 α}$ is

x^{2} + y^{2} + z^{2}

4 x^{2} - 5 y^{2} + 3 z^{2}

- 3 x^{2} + 4 y^{2} + 2 z^{2}

Express each of the following as an algebraic sum of sines or cosines :

(i) 2 sin $3 x$ cos $2 x$

(ii) 2 cos $4 x$ sin 2 $x$

(iii) 2 cos $6 x$ cos $4 x$

(iv) 2 sin $3 x$ sin $5 x$

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