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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
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