Question

Prove that :-
$\rm \cos6x = 32 {cos}^{6}x - 48 {cos}^{4}x + 18 { \cos }^{2}x - 1$
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Answer 1

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

Consider

$\rm \: cos6x$

can be rewritten as

$\rm \: = \: cos2(3x)$

We know,

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

So, using this identity, we get

$\rm \: = \: {2cos}^{2}3x - 1$

can be further rewritten as

$\rm \: = \: 2 {(cos3x)}^{2} - 1$

We know,

$\boxed{\tt{ cos3x = {4cos}^{3}x - 3cosx \: }}$

So, using this identity, we get

$\rm \: = \: 2 {(4 {cos}^{3} x - 3cosx)}^{2} - 1$

We know,

$\boxed{\tt{ {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy \: }}$

So, using this identity, we get

$\rm \: = \: 2\bigg[ \: 16 {cos}^{6}x + {9cos}^{2}x - 24 {cos}^{4}x \: \bigg] - 1$

$\rm \: = \:\: 32 {cos}^{6}x + {18cos}^{2}x - 48 {cos}^{4}x \: - 1$

can be re-arranged as

$\rm \: = \:\: 32 {cos}^{6}x \: - \: {48cos}^{4}x \: + \: 18{cos}^{2}x \: - 1$

Hence,

$\boxed{\tt{ \cos6x = 32 {cos}^{6}x - 48 {cos}^{4}x + 18 { \cos }^{2}x - 1 \: }}$

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ADDITIONAL INFORMATION

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

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

$\boxed{\tt{ cos2x = \frac{1 - {tan}^{2} x}{1 + {tan}^{2}x } }}$

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

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

$\boxed{\tt{ sin2x = 2 \: sinx \: cosx \: }}$

$\boxed{\tt{ sin3x \: = \: 3sinx \: - \: {4sin}^{3}x \: }}$

Answer 2

Question :-

Prove that :

$\mapsto \bf cos6x =\: 32cos^{6}x - 48cos^{4}x + 18cos^{2}x - 1$

Solution :-

$\bigstar\: \: \sf\bold{\purple{L.H.S =\: cos6x}}$

$\implies \sf cos2.(3x)$

As we know that :

$\clubsuit\: \: \sf\bold{\pink{cos2x =\: 2cos^{2}x - 1}}$

$\clubsuit\: \: \sf\bold{\pink{cos 3x =\: 4cos^{3}x - 3 cosx}}$

So, by replacing by 3x we get,

$\implies \sf cos2(3x) =\: 2cos^{2}(3x) - 1$

$\implies \sf cos6x =\: 2cos^{2} 3x - 1$

$\implies \sf 2(cos 3x)^2 - 1$

Now, by putting cos 3 x = 4 cos³ x - 3 cos x we get,

$\implies \sf 2(4 cos^{3}x - 3 cos x)^2 - 1$

Again, as we know that :

$\clubsuit\: \: \sf\bold{\pink{(a - b)^2 =\: a^2 + b^2 - 2ab}}$

So, by using this identities we get,

$\implies \sf 2\bigg[(4 cos^3 x)^2 + (3 cos x)^2 - 2(4 cos^{3} x) \times (3 cos x)\bigg] - 1$

$\implies \sf 2\bigg[(16 cos^{6} x + 9 cos^{2} x - 24 cos^{4} x)\bigg] - 1$

$\implies \sf 2 \times 16 cos^{6} x + 2 \times 9 cos^{2} x - 2 \times 24 cos^{4} x - 1$

$\implies \sf 32 cos^{6}x + 18 cos^{2}x - 48 cos^{4} x - 1$

By rearrange we get,

$\implies \sf\bold{\red{32 cos^{6}x - 48 cos^{4} x + 18 cos^{2} x - 1}}$

Again,

$\bigstar\: \: \sf\bold{\purple{R.H.S =\: 32 cos^{6} x - 48 cos^{4} x + 18 cos^{2} x - 1}}$

$\implies \sf\bold{\red{32 cos^{6}x - 48 cos^{4} x + 18 cos^{2} x - 1}}$

$\leadsto \sf\bold{L.H.S =\: R.H.S}$

$\: \: \: \: \: \: \: \: \: \: \: \: \sf\boxed{\bold{\green{HENCE\: PROVED}}}\: \:$

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