Proof of Compound Angle Formula
cos\(^{2}\) α - sin\(^{2}\) β

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

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

Proof: cos (α + β) cos (α - β)

= (cos α cos β - sin α sin β) (cos α cos β + sin α sin β)

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

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

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

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

= cos\(^{2}\) α - sin\(^{2}\) β

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

= 1 - sin\(^{2}\) α - 1 + cos\(^{2}\) β

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

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


Solved examples using the proof of compound angle formula cos\(^{2}\)α - sin\(^{2}\) β:

1. Prove that: cos\(^{2}\) 2x - sin\(^{2}\) x = cos x cos 3x.

Solution:

L.H.S. = cos\(^{2}\) 2x - sin\(^{2}\) x

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

= cos 3x cos x = R.H.S.                         Proved

 

2. Find the value of cos\(^{2}\) (\(\frac{π}{8}\) - \(\frac{θ}{2}\)) - sin\(^{2}\) (\(\frac{π}{8}\) + \(\frac{θ}{2}\)).

Solution:

cos\(^{2}\) (\(\frac{π}{8}\) - \(\frac{θ}{2}\)) - sin\(^{2}\) (\(\frac{π}{8}\) + \(\frac{θ}{2}\))

= cos {(\(\frac{π}{8}\) - \(\frac{θ}{2}\)) + (\(\frac{π}{8}\) + \(\frac{θ}{2}\))} cos {(\(\frac{π}{8}\) - \(\frac{θ}{2}\)) - (\(\frac{π}{8}\) + \(\frac{θ}{2}\))},

[since we know, cos\(^{2}\) α - sin\(^{2}\) β = cos (α + β)

cos (α - β)]

= cos {\(\frac{π}{8}\) - \(\frac{θ}{2}\) + \(\frac{π}{8}\) + \(\frac{θ}{2}\)} cos {\(\frac{π}{8}\) - \(\frac{θ}{2}\) - \(\frac{π}{8}\) - \(\frac{θ}{2}\)}

= cos {\(\frac{π}{8}\) + \(\frac{π}{8}\)} cos {- \(\frac{θ}{2}\) - \(\frac{θ}{2}\)}

= cos \(\frac{π}{4}\) cos (- θ)

= cos \(\frac{π}{4}\) cos θ, [since we know, cos (- θ) = cos θ)

= \(\frac{1}{√2}\) ∙ cos θ [we know, cos \(\frac{π}{4}\) = \(\frac{1}{√2}\)]

 

3. Evaluate: cos\(^{2}\) (\(\frac{π}{4}\) + x) - sin\(^{2}\) (\(\frac{π}{4}\) - x)

Solution:

cos\(^{2}\) (\(\frac{π}{4}\) + x) - sin\(^{2}\) (\(\frac{π}{4}\) - x)

= cos {(\(\frac{π}{4}\) + x) + (\(\frac{π}{4}\) - x)} cos {(\(\frac{π}{4}\) + x) - (\(\frac{π}{4}\) - x)}, [since we know, cos\(^{2}\) β - sin\(^{2}\) α = cos (α + β)

cos (α - β)]

= cos {\(\frac{π}{4}\) + x + \(\frac{π}{4}\) - x} cos {\(\frac{π}{4}\) + x - \(\frac{π}{4}\) + x}

= cos {\(\frac{π}{4}\)+\(\frac{π}{4}\)} cos {x + x}

= cos \(\frac{π}{4}\) cos 2x

= 0 ∙ cos 2x, [Since we know, cos \(\frac{π}{4}\) = 0]

= 0






11 and 12 Grade Math

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