If cos 3 x -sin 2 x-x-0n arsin - x- - x - R- then

The correct options are B n=5, a_1=1/4 C n=5, a_3=1/8 cos^3 x .sin 2x=cos^2x. cosx.sin 2x ∑_x=0^na_r sin(rx)= ( 1 cos 2x/2 ) ( 2 sin 2x cos x/2 ) =1/4(1 cos 2 ...

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Byju's Answer

Standard XI

Mathematics

Product of Trigonometric Ratios in Terms of Their Sum

If cos 3 x ·s...

Question

If $c o s^{3} x . s i n 2 x = \sum_{x = 0}^{n} a_{r} s i n (π x), \forall x \in R$ , then

$n = 5, a_{1} = \frac{1}{2}$

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$n = 5, a_{1} = \frac{1}{4}$

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$n = 5, a_{3} = \frac{1}{8}$

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$n = 5, a_{5} = \frac{1}{4}$

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Solution

The correct options are
B
$n = 5, a_{1} = \frac{1}{4}$

C
$n = 5, a_{3} = \frac{1}{8}$

$c o s^{3} x . s i n 2 x = c o s^{2} x . c o s x . s i n 2 x \sum_{x = 0}^{n} a_{r} s i n (r x) = (\frac{1 - c o s 2 x}{2}) (\frac{2 s i n 2 x c o s x}{2}) = \frac{1}{4} (1 - c o s 2 x) (s i n 3 x + s i n x) = \frac{1}{4} (s i n 3 x + s i n x - \frac{1}{2} (2 s i n 3 x c o s 2 x) - \frac{1}{2} (2 c o s 2 x s i n x)) = \frac{1}{4} (s i n 3 x + s i n x - \frac{1}{2} (s i n 5 x + s i n x) - \frac{1}{2} (s i n 3 x - s i n x)) \sum_{x = 0}^{n} a_{r} (s i n r x) = \frac{1}{4} (s i n x + \frac{1}{2} s i n 3 x - \frac{1}{2} s i n 5 x)$
$∴ a_{0} s i n 0 x + a_{1} s i n x + a_{2} s i n 2 x + a_{3} s i n 3 x + a_{4} s i n 4 x + a_{5} s i n 5 x = \frac{1}{4} (s i n x) + \frac{1}{8} s i n 3 x - \frac{1}{8} s i n 5 x .$
$n = 5.$
$a_{1} = \frac{1}{4}, a_{3} = \frac{1}{8}, a_{5} = - \frac{1}{8}$

Suggest Corrections

If cos3x.sin2x=∑nx=0arsin(πx),∀x∈R, then

If $c o s^{3} x . s i n 2 x = \sum_{x = 0}^{n} a_{r} s i n (π x), \forall x \in R$ , then