# Proof of Compound Angle Formula sin$$^{2}$$ α - sin$$^{2}$$ β

We will learn step-by-step the proof of compound angle formula sin$$^{2}$$ α - sin$$^{2}$$ β. We need to take the help of the formula of sin (α + β) and sin (α - β) to proof the formula of sin$$^{2}$$ α - sin$$^{2}$$ β for any positive or negative values of α and β.

Prove that sin (α + β) sin (α - β) = sin$$^{2}$$ α - sin$$^{2}$$ β = cos$$^{2}$$ β - cos$$^{2}$$ α.

Proof: sin(α + β) sin (α + β)

= (sin α cos β + cos α sin β) (sin α cos β - cos α sin β); [applying the formula of sin (α + β) and sin (α - β)]

= (sin α cos β)$$^{2}$$ - (cos α sin β)$$^{2}$$

= sin$$^{2}$$ α cos$$^{2}$$ β - cos$$^{2}$$ α sin$$^{2}$$ β

= sin$$^{2}$$ α (1 - sin$$^{2}$$ β) - (1 - sin$$^{2}$$ α) sin$$^{2}$$ β; [since we know, cos$$^{2}$$ θ = 1 - sin$$^{2}$$ θ]

= sin$$^{2}$$ α - sin$$^{2}$$ α sin$$^{2}$$ β - sin$$^{2}$$ β + sin$$^{2}$$ α sin$$^{2}$$ β

= sin$$^{2}$$ α - sin$$^{2}$$ β

= 1 - cos$$^{2}$$ α - (1 - cos$$^{2}$$ β); [since we know, sin$$^{2}$$ θ = 1 - cos$$^{2}$$ θ]

= 1 - cos$$^{2}$$ α - 1 + cos$$^{2}$$ β

= cos$$^{2}$$ β - cos$$^{2}$$ α                     Proved

Therefore, sin (α + β) sin (α - β) = sin$$^{2}$$ α - sin$$^{2}$$ β = cos$$^{2}$$ β - cos$$^{2}$$ α

Solved examples using the proof of compound angle formula sin$$^{2}$$ α - sin$$^{2}$$ β:

1. Prove that sin$$^{2}$$ 6x - sin$$^{2}$$ 4x = sin 2x sin 10x.

Solution:

L.H.S. = sin$$^{2}$$ 6x - sin$$^{2}$$ 4x

= sin (6x + 4x) sin (6x - 4x); [since we know sin$$^{2}$$ α - sin$$^{2}$$ β = sin (α + β) sin (α - β)]

= sin 10x sin 2x = R.H.S.                         Proved

2. Prove that cos$$^{2}$$ 2x - cos$$^{2}$$ 6x = sin 4x sin 8x.

Solution:

L.H.S. = cos$$^{2}$$ 2x - cos$$^{2}$$ 6x

= (1 - sin$$^{2}$$ 2x) - (1 - sin$$^{2}$$ 6x), [since we know cos$$^{2}$$ θ = 1 - sin$$^{2}$$ θ]

= 1 - sin$$^{2}$$ 2x - 1 + sin$$^{2}$$ 6x

= sin$$^{2}$$ 6x - sin$$^{2}$$ 2x

= sin (6x + 2x) sin (6x - 2x), [since we know sin$$^{2}$$ α - sin$$^{2}$$ β = sin (α + β) sin (α - β)]

= sin 8x sin 4x = R.H.S.                         Proved

3. Evaluate: sin$$^{2}$$ ($$\frac{π}{8}$$ + $$\frac{x}{2}$$) - sin$$^{2}$$ ($$\frac{π}{8}$$ - $$\frac{x}{2}$$).

Solution:

sin$$^{2}$$ ($$\frac{π}{8}$$ + $$\frac{x}{2}$$) - sin$$^{2}$$ ($$\frac{π}{8}$$ - $$\frac{x}{2}$$)

= sin {($$\frac{π}{8}$$ + $$\frac{x}{2}$$) + ($$\frac{π}{8}$$ - $$\frac{x}{2}$$)} sin {($$\frac{π}{8}$$ + $$\frac{x}{2}$$) - ($$\frac{π}{8}$$ - $$\frac{x}{2}$$)}, [since we know sin$$^{2}$$ α - sin$$^{2}$$ β = sin (α + β) sin (α - β)]

= sin {$$\frac{π}{8}$$ + $$\frac{x}{2}$$ + $$\frac{π}{8}$$ - $$\frac{x}{2}$$} sin {$$\frac{π}{8}$$ + $$\frac{x}{2}$$ - $$\frac{π}{8}$$ + $$\frac{x}{2}$$}

= sin {$$\frac{π}{8}$$ + $$\frac{π}{8}$$} sin {$$\frac{x}{2}$$ + $$\frac{x}{2}$$}

= sin $$\frac{π}{4}$$ sin x

= $$\frac{1}{√2}$$ sin x, [Since we know sin $$\frac{π}{4}$$ = $$\frac{1}{√2}$$]

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