Question

Prove that

$sin2x + 2sin4x + sin6x = {4cos}^{2}x \: sin4x$
Step by step explanation​

Answer 1

$\large\underline{\sf{Solution-}}$

Consider LHS

$\rm \: sin2x + 2sin4x + sin6x$

can be re-arranged as

$\rm \: = \: (sin6x + sin2x) + 2sin4x$

We know,

$\boxed{\tt{ sinx + siny = 2sin\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg] \: }}$

So, on substituting the values, we get

$\rm \: = \: 2sin\bigg[\dfrac{6x + 2x}{2} \bigg]cos\bigg[\dfrac{6x - 2x}{2} \bigg] + 2sin4x$

$\rm \: = \: 2sin\bigg[\dfrac{8x}{2} \bigg]cos\bigg[\dfrac{4x}{2} \bigg] + 2sin4x$

$\rm \: = \: 2sin4x \: cos2x \: + \: 2sin4x$

$\rm \: = \: 2sin4x(cos2x + 1)$

We know,

$\boxed{\tt{ cos2x = {2cos}^{2}x - 1 \: }}$

So, using this identity, we get

$\rm \: = \: 2sin4x( {2cos}^{2}x - 1 + 1)$

$\rm \: = \: 2sin4x( {2cos}^{2}x)$

$\rm \: = \: {4cos}^{2}x \: sin4x$

Hence,

$\rm\implies \:\boxed{\tt{ \rm \:sin2x + 2sin4x + sin6x={4cos}^{2}x sin4x}}$

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ADDITIONAL INFORMATION

$\boxed{\tt{ sinx - siny = 2cos\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg] \: }}$

$\boxed{\tt{ cosx + cosy = 2cos\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg] \: }}$

$\boxed{\tt{ cosx - cosy = - 2sin\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg] \: }}$

$\boxed{\tt{ 2sinx \: cosy \: = \: sin(x + y) + sin(x - y) \: }}$

$\boxed{\tt{ 2cosx \: cosy \: = \: cos(x + y) + cos(x - y) \: }}$

$\boxed{\tt{ 2sinx \: siny \: = \: cos(x - y) - cos(x + y) \: }}$

$\boxed{\tt{ sin2x = 2sinx \: cosx \: = \: \frac{2tanx}{1 + {tan}^{2} x} \: }}$

$\boxed{\tt{ cos2x = {cos}^{2}x - {sin}^{2}x = 1 - {2sin}^{2}x \: }}$

$\boxed{\tt{ tan2x = \frac{2tanx}{1 - {tan}^{2}x } \: }}$

Answer 2

$\large{ \underline{ \sf \: solution}}$

LHS

= sin2x+2sin4x+sin 6x

Re-arranged

$= (\sin6x </u><u>+</u><u>\sin2x) + 2\sin 4x$

So,

sin x + sin y = 2sin [x+y/2] cos[x-y/2]

The e substituting value we get,

2sin[6x+2x/2] cos [6x-2x/2]+2sin4x

2sin[8x/2] cos [6x-2x/2]+sin 4x

2sin4x cos 2x+2sin4x

2sin4x(cos 2x+1)

Now,

cos 2x = 2cos² x - 1

Using identity,

2sin4x(2cos² x - 1 + 1)

2sin4x(2cos²x)

4cos²x sin 4x

Hence,

(sin 2x+2sin4x+sin 6x = 4cos² x sin 4x)