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Proof of Compound Angle Formula
cos2 α - sin2 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) = cos2 α - sin2 β = cos2 β - sin2 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

= cos2 α cos2 β - sin2 α sin2 β

= cos2 α (1 - sin2 β) - (1 - cos2 α) sin2 β, [since we know, cos2 θ = 1 - sin2 θ]

= cos2 α - cos2 α sin2 β - sin2 β + cos2 α sin2 β

= cos2 α - sin2 β

= 1 - sin2 α - (1 - cos2 β), [since we know, cos2 θ = 1 - sin2 θ and sin2 θ = 1 - cos2 θ]

= 1 - sin2 α - 1 + cos2 β

= cos2 β - sin2 α                     Proved

Therefore, cos (α + β) cos (α - β) = cos2 α - sin2 β = cos2 β - sin2 α


Solved examples using the proof of compound angle formula cos2α - sin2 β:

1. Prove that: cos2 2x - sin2 x = cos x cos 3x.

Solution:

L.H.S. = cos2 2x - sin2 x

= cos (2x + x) cos (2x - x), [since we know cos2 α - sin2 β = cos (α + β) cos (α - β)]

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

 

2. Find the value of cos2 (\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

 Compound Angle






11 and 12 Grade Math

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