Question

prove that: sin(π/4+x)*sin(π/4-x)=1/2cos2x​

Answer 1

Answer:

answer for the given problem is given

Answer 2

Answer- The above question is from the chapter 'Trigonometric functions'.

Trigonometry- The branch of Mathematics which helps in dealing with measure of three sides of a right-angled triangle is called Trigonometry.

Trigonometric Ratios:

sin θ  = Perpendicular/Hypotenuse

cos θ = Base/Hypotenuse

tan θ = Perpendicular/Base

cosec θ = Hypotenuse/Perpendicular

sec θ = Hypotenuse/Base

cot θ = Base/Perpendicular

Also, tan θ = sin θ/cos θ and cot θ = cos θ/sinθ

Concept used: 1) sin (A + B) = sinAcosB + sinBcosA

2) sin (A - B) = sinAcosB - sinBcosA

3) cos² x - sin² x = cos 2x

Given question: Prove that-

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

Solution:

$L.HS. = sin \: (\dfrac{\pi}{4} + x) \: sin \: (\dfrac{\pi}{4} - x)\\\\\\= [sin \dfrac{\pi}{4}\: cos \: x + sin \: x \: cos \dfrac{\pi}{4}] . [sin \dfrac{\pi}{4}\: cos \: x + sin \: x \: cos \dfrac{\pi}{4}]\\\\= [\dfrac{1}{\sqrt{2}}\: cos \: x + sin \: x \: \dfrac{1}{\sqrt{2}}] [\dfrac{1}{\sqrt{2}}\: cos \: x - sin \: x \: \dfrac{1}{\sqrt{2}}] \\\\= [\dfrac{1}{\sqrt{2}} (cos \:x \: + sin \: x)] [\dfrac{1}{\sqrt{2}} (cos \:x \: - sin \: x)]\\\\\\= \dfrac{1}{2} (cos^{2} \: x \: - sin^{2} \: x)$

$= \dfrac{1}{2} \: cos \: 2x\\\\= R.H.S.\\\\Hence, proved.$