Question

Prove that sin square 75 - cos square 45 is equals to root 3 by 4

Answer 1

Question :-

Prove that

$\rm \: {sin}^{2}75\degree - {cos}^{2}45\degree = \dfrac{ \sqrt{3} }{4}$

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

Consider LHS

$\rm \: {sin}^{2}75\degree - {cos}^{2}45\degree$

can be rewritten as

$\rm \: = \: - ( {cos}^{2}45\degree - {sin}^{2}75\degree )$

We know,

$\boxed{\sf{ \: {cos}^{2}x - {sin}^{2}y = cos(x + y) \: cos(x - y) \: }}$

So, using this, we get

$\rm \: = \: - \bigg(cos(45\degree + 75\degree ) \: cos(45\degree - 75\degree) \bigg)$

$\rm \: = \: - \bigg(cos(120\degree) \: cos( - 30\degree) \bigg)$

We know,

$\boxed{\sf{ \:cos( - x) = cosx \: }}$

So, using this, we get

$\rm \: = \: - cos(180\degree - 60\degree ) \: cos30\degree$

We know,

$\boxed{\sf{ \:cos(180\degree - x) \: = \: - \: cosx \: }}$

So, using this, we get

$\rm \: = \: cos60\degree \: cos30\degree$

$\rm \: = \: \dfrac{ \sqrt{3} }{2} \times \dfrac{1}{2}$

$\rm \: = \: \dfrac{ \sqrt{3} }{4}$

Hence,

$\rm\implies \:\boxed{\sf{ \: \: \rm \: {sin}^{2}75\degree - {cos}^{2}45\degree = \dfrac{ \sqrt{3} }{4} \: }}$

$\green{\large\underline{\sf{Aliter \: Solution-}}}$

Consider LHS

$\rm \: {sin}^{2}75\degree - {cos}^{2}45\degree$

can be rewritten as

$\rm \: \rm \: \dfrac{1}{2}\bigg(2{sin}^{2}75\degree - 2{cos}^{2}45\degree\bigg)$

We know,

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

and

$\boxed{\sf{ \:cos45\degree = \frac{1}{ \sqrt{2} } \: }}$

So, using this, we get

$\rm \: = \: \dfrac{1}{2}\bigg(1 - cos150\degree - 2 \times \dfrac{1}{2} \bigg)$

$\rm \: = \: \dfrac{1}{2}\bigg(1 - cos(180\degree - 30\degree ) - 1 \bigg)$

We know,

So, using this, we get

$\rm \: = \: \dfrac{1}{2}\bigg(cos30\degree\bigg)$

$\rm \: = \: \dfrac{1}{2} \times \dfrac{ \sqrt{3} }{2}$

$\rm \: = \: \dfrac{ \sqrt{3} }{4}$

Hence,

$\rm\implies \:\boxed{\sf{ \: \: \rm \: {sin}^{2}75\degree - {cos}^{2}45\degree = \dfrac{ \sqrt{3} }{4} \: }}$

$\green{\large\underline{\sf{Aliter \: Solution-}}}$

Consider LHS

$\rm \: {sin}^{2}75\degree - {cos}^{2}45\degree$

Consider

$\rm \: sin75\degree$

$\rm \: = \: sin(45\degree + 30\degree )$

$\rm \: = \: sin45\degree cos30\degree + sin30\degree cos45\degree$

$\rm \: = \: \dfrac{1}{ \sqrt{2} } \times \dfrac{ \sqrt{3} }{2} + \dfrac{1}{ \sqrt{2} } \times \dfrac{1}{2}$

$\rm \: = \: \dfrac{ \sqrt{3} }{2 \sqrt{2} } + \dfrac{1}{2 \sqrt{2} }$

$\rm \: = \: \dfrac{ \sqrt{3} + 1 }{2 \sqrt{2} }$

$\rm\implies \:sin75\degree = \: \dfrac{ \sqrt{3} + 1 }{2 \sqrt{2} }$

Now, Consider

$\rm \: {sin}^{2}75\degree - {cos}^{2}45\degree$

So, on substituting the values, we get

$\rm \: = \: {\bigg(\dfrac{ \sqrt{3} + 1 }{2 \sqrt{2} } \bigg) }^{2} - {\bigg(\dfrac{1}{ \sqrt{2} } \bigg) }^{2}$

$\rm \: = \: \dfrac{3 + 1 + 2 \sqrt{3} }{8} - \dfrac{1}{2}$

$\rm \: = \: \dfrac{4+ 2 \sqrt{3} }{8} - \dfrac{1}{2}$

$\rm \: = \: \dfrac{4+ 2 \sqrt{3} - 4}{8}$

$\rm \: = \: \dfrac{2 \sqrt{3}}{8}$

$\rm \: = \: \dfrac{\sqrt{3}}{4}$

Hence,

$\rm\implies \:\boxed{\sf{ \: \: \rm \: {sin}^{2}75\degree - {cos}^{2}45\degree = \dfrac{ \sqrt{3} }{4} \: }}$

Answer 2

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