If sin3 x sin 3x-m-0ncm cos mx where c0- c1- c2- - - cn are constants and cn - 0- then the value of n is

Sin^3 x sin 3x=1/4(3 sin x sin 3x)sin 3x =3/8.2 sin x sin 3x 1/8.2 sin^2 3x =3/8(cos 2x cos 4x) 1/8(1 cos 6x) = 1/8+3/8 cos 2x 3/8cos 4x+1/8cos 6x ....(i) ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Multiples of an Angle

If sin3 x sin...

Question

$I f s i n^{3} x s i n 3 x = \sum_{m = 0}^{n} c_{m} c o s m x w h e r e c_{0}, c_{1}, c_{2} . \dots \dots, c_{n} a r e c o n s t a n t s a n d c_{n} \neq 0, t h e n t h e v a l u e o f n i s$

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Solution

The correct option is B
6

$s i n^{3} x s i n 3 x = \frac{1}{4} (3 s i n x - s i n 3 x) s i n 3 x = \frac{3}{8} .2 s i n x s i n 3 x - \frac{1}{8} .2 s i n^{2} 3 x = \frac{3}{8} (c o s 2 x - c o s 4 x) - \frac{1}{8} (1 - c o s 6 x) = - \frac{1}{8} + \frac{3}{8} c o s 2 x - \frac{3}{8} c o s 4 x + \frac{1}{8} c o s 6 x . . . . (i)$

$a n d \sum_{m = 0}^{n} c_{m} c o s m x = c_{0} + c_{1} c o s x + c_{2} c o s 2 x + c_{3} c o s 3 x + . . . . . + c_{n} c o s n x . . . . . . (i i)$

Comparing both sides of (i) and (ii), we get n = 6

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If sin3 x sin 3x=∑nm=0cm cos mx where c0,c1,c2.⋯⋯,cn are constants and cn≠0, then the value of n is

$I f s i n^{3} x s i n 3 x = \sum_{m = 0}^{n} c_{m} c o s m x w h e r e c_{0}, c_{1}, c_{2} . \dots \dots, c_{n} a r e c o n s t a n t s a n d c_{n} \neq 0, t h e n t h e v a l u e o f n i s$