Learn Prove That Sin C D X Sin C D Sin2c Sin2d

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# Prove that: sin (C + D) × sin (C-D) = sin^{2}C - sin^{2}D

Trigonometric Ratios is a branch of mathematics that deals with the relation between the angles and sides of a triangle.

## Answer: sin(C+D) sin(C-D) = sin^{2 }C – sin^{2} D

Let see, how we can solve

**Explanation:**

Let's look into the following formulas

sin(C+D)=sin(C)cos(D)+cos(C)sin(D)

sin(C−D) = sin(C)cos(D)−cos(C)sin(D)

Therefore,

LHS = sin(C+D) × sin(C−D)

= (sinC cosD+cosC sinD) × (sinC cosD−cosC sinD)

= (sinC cosD)^{2}−(cosC sinD)^{2} (By using the identity (c+d)(c−d)=c^{2} – d^{2})

= sin^{2}C cos^{2}D−sin^{2}D cos^{2}C

= sin^{2}C(1−sin^{2}D)−sin^{2}D(1−sin^{2}C) (Since, sin^{2}θ+cos^{2}θ=1, thus cos^{2}θ = 1 - sin^{2}θ)

= sin^{2}C−sin^{2}D−sin^{2}C. sin^{2}D+sin^{2}D sin^{2}C

= sin^{2}C−sin^{2}D = RHS

### Thus, verified sin(C+D) × sin(C-D) = sin^{2} C – sin^{2} D

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