Question

Answer the 31th question rightly and I'll mark it as brainliest.​

Answer 1

Required Answer:-

Given to prove:

$\sf \mapsto \cos(x) \cos(2x) \cos(4x) \cos(8x) = \dfrac{ \sin(16x) }{16 \sin(x) }$

Proof:

Taking LHS,

$\sf\cos(x) \cos(2x) \cos(4x) \cos(8x)$

$\sf = \dfrac{2 \sin(x) }{2 \sin(x) } \{\cos(x) \}\cos(2x) \cos(4x) \cos(8x)$

$\sf = \dfrac{2 \sin(x) \cos(x) }{2 \sin(x) } \big\{ \cos(2x) \cos(4x) \cos(8x) \big\}$

Now, remember this formula.

$\sf \mapsto 2 \sin( \alpha ) \cos( \alpha ) = \sin(2 \alpha )$

Using this formula, we get,

$\sf = \dfrac{\sin(2x) }{2 \sin(x) } \big\{ \cos(2x) \cos(4x) \cos(8x) \big\}$

$\sf = \dfrac{\sin(2x) \cos(2x) }{2 \sin(x) } \big\{\cos(4x) \cos(8x) \big\}$

Multiplying numerator and denominator by 2, we get,

$\sf = \dfrac{2 \times \sin(2x) \cos(2x) }{2 \times 2 \sin(x) } \big\{\cos(4x) \cos(8x) \big\}$

Now, again apply the same formula.

$\sf = \dfrac{\sin(2 \times 2x) }{4 \sin(x) } \big\{\cos(4x) \cos(8x) \big\}$

$\sf = \dfrac{\sin(4x) \cos(4x) }{4 \sin(x) } \big\{\cos(8x) \big\}$

Again, multiplying both numerator and denominator by 2,we get,

$\sf = \dfrac{2 \times \sin(4x) \cos(4x) }{2 \times 4 \sin(x) } \big\{\cos(8x) \big\}$

$\sf = \dfrac{2\sin(4x) \cos(4x) }{8 \sin(x) } \big\{\cos(8x) \big\}$

Again, applying the same formula, we get,

$\sf = \dfrac{\sin(2 \times 4x) }{8 \sin(x) } \big\{\cos(8x) \big\}$

$\sf = \dfrac{\sin(8x) \cos(8x) }{8 \sin(x) }$

Multiplying both numerator and denominator by 2, we get,

$\sf = \dfrac{2 \times \sin(8x) \cos(8x) }{2 \times 8 \sin(x) }$

$\sf = \dfrac{2\sin(8x) \cos(8x) }{16 \sin(x) }$

Applying the same formula, we get,

$\sf = \dfrac{\sin(2 \times 8x) }{16 \sin(x) }$

$\sf = \dfrac{\sin(16x) }{16 \sin(x) }$

Taking RHS,

$\sf = \dfrac{\sin(16x) }{16 \sin(x) }$

Hence, LHS = RHS (Proved)