Question

Prove that's cos square A +cos square (60+A) +cos square (60-A) =3/2

Answer 1

Question :-

Prove that

$\rm \: {cos}^{2}A + {cos}^{2}(60\degree + A) + {cos}^{2}(60\degree - A) = \dfrac{3}{2}$

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

Consider LHS

$\rm \: {cos}^{2}A + {cos}^{2}(60\degree + A) + {cos}^{2}(60\degree - A)$

can be rewritten as

$\rm \: = \: \dfrac{1}{2}\bigg(2{cos}^{2}A + 2{cos}^{2}(60\degree + A) + 2{cos}^{2}(60\degree - A)\bigg)$

We know,

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

So, using this identity, we get

$\rm \: = \: \dfrac{1}{2} \bigg(1 + cos2A + 1 + cos(120\degree + 2A) + 1 + cos(120\degree - 2A)\bigg)$

$\rm \: = \: \dfrac{1}{2} \bigg(3 + cos2A + cos(120\degree + 2A) + cos(120\degree - 2A)\bigg)$

We know,

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

So, using this identity, we get

$\rm \: = \: \dfrac{1}{2} \bigg(3 + cos2A + 2cos120\degree cos2A\bigg)$

can be rewritten as

$\rm \: = \: \dfrac{1}{2} \bigg(3 + cos2A + 2cos(180\degree - 60\degree ) cos2A\bigg)$

We know,

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

So, using this result, we get

$\rm \: = \: \dfrac{1}{2} \bigg(3 + cos2A - 2cos60\degree \: cos2A\bigg)$

$\rm \: = \: \dfrac{1}{2} \bigg(3 + cos2A - 2 \times \dfrac{1}{2} \times \: cos2A\bigg)$

$\rm \: = \: \dfrac{1}{2} \bigg(3 + cos2A - \: cos2A\bigg)$

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

Hence,

$\rm\implies \:\boxed{\sf{ \:\rm \: {cos}^{2}A + {cos}^{2}(60\degree + A) + {cos}^{2}(60\degree - A) = \frac{3}{2} \: }}$

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Additional Information :-

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

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

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

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

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

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

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

Answer 2

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