Question

Sin(π/4+x)sin(π/4-x)=1/2cos2x


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Answer 1

Answer:

L.H.S. = R.H.S.

Step-by-step explanation:

Given :

$\large sin(\dfrac{\pi}{4}+x) \ sin(\dfrac{\pi}{4}-x)=\dfrac{1}{2} \ cos \ 2x$

$\large L.H.S.=sin(\dfrac{\pi}{4}+x) \ sin(\dfrac{\pi}{4}-x)\\\\\\\large L.H.S.=sin(x+\dfrac{\pi}{4}) \ sin(-x+\dfrac{\pi}{4})\\\\\\\large \text{Using formula sin(a+\dfrac{b}{2})sin(a+\dfrac{b}{2})=-\dfrac{1}{2}(cos \ a-cos \ b) }\\\\\\\large L.H.S.=-\dfrac{1}{2}(cos \ \dfrac{2\pi}{4} -cos \ 2x)\\\\\\\large \text{we know that cos \ \dfrac{\pi}{2}=0 }$

$\large L.H.S.=-\dfrac{1}{2}(0-cos \ 2x)\\\\\\\large L.H.S.=-\dfrac{1}{2}(-cos \ 2x)\\\\\\\large L.H.S.=\dfrac{1}{2}(cos \ 2x)$

L.H.S. = R.H.S.

Hence proved.

Thus we get answer.

Answer 2

Step-by-step explanation:

Sin(π/4+x)sin(π/4-x)

=1/2(2Sin(π/4+x)sin(π/4-x))

=1/2*[cos(π/4-x-π/4-x)/2+cos(π/4-x+π/4+x)

=1/2 [cos(-2x)+cos(π/4+π/4)]

=1/2 [cos2x+0]

=1/2 cos2x