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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

● Compound Angle