Question

sin (3π/4 +x) + sin (3π/4-x) =√2 cos x​

Answer 1

To prove :

$\red{\bullet}\sf{\;sin\left(\dfrac{3\;\pi}{4}+x\right)+sin\left(\dfrac{3\;\pi}{4}-x\right)=\sqrt{2}\;cos \;x}$

Proof :

Taking LHS

$\implies\sf{LHS=\;sin\left(\dfrac{3\;\pi}{4}+x\right)+sin\left(\dfrac{3\;\pi}{4}-x\right)}$

Using trigonometric identities

sin (x + y) = sin x cos y + cos x sin y &

sin ( x - y) = sin x cos y - cos x sin y

$\implies\sf{LHS=sin\dfrac{3\;\pi}{4}\;cos\;x + cos \dfrac{3\pi}{4}\;sin\;x+\left(sin\dfrac{3\;\pi}{4}\;cos\;x - cos \dfrac{3\pi}{4}\;sin\;x\right) }$

$\implies\sf{LHS=sin\dfrac{3\;\pi}{4}\;cos\;x + cos \dfrac{3\pi}{4}\;sin\;x+sin\dfrac{3\;\pi}{4}\;cos\;x - cos \dfrac{3\pi}{4}\;sin\;x}$

$\implies\sf{LHS=sin\dfrac{3\;\pi}{4}\;cos\;x+sin\dfrac{3\;\pi}{4}\;cos\;x }$

$\implies\sf{LHS=2\;sin\dfrac{3\;\pi}{4}\;cos\;x }$

putting value of  $\bf{sin\;\dfrac{3\;\pi}{4}}$ = $\bf{\dfrac{1}{\sqrt{2}}}$

$\implies\sf{LHS=2\;\big(\dfrac{1}{\sqrt{2}}\big)\;cos\;x }$

$\implies\sf{LHS=\sqrt{2}\;cos\;x = RHS }$

PROVED.

Few more related Identities :

• cos ( x + y ) = cos x cos y - sin x sin y
• cos ( x - y ) = cos x cos y + sin x sin y
• sin ( -x ) = - sin x
• cos ( -x ) = cos x
• cos 2 x = cos²x - sin²x
• sin 2 x = 2 sin x cos x