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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