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 \(\frac{π}{4}\) = \(\frac{1}{√2}\)]






11 and 12 Grade Math

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