Prove that: sin (C + D) × sin (C-D) = sin²C - sin²D

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) = sin² C – sin² 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)=c² – d²) 

= sin²C cos²D−sin²D cos²C 

= sin²C(1−sin²D)−sin²D(1−sin²C)                    (Since, sin²θ+cos²θ=1, thus cos²θ = 1 -  sin²θ)

= sin²C−sin²D−sin²C. sin²D+sin²D sin²C

= sin²C−sin²D = RHS

Thus, verified sin(C+D) × sin(C-D) = sin² C – sin² D