# 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

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