# Proof cos(x+y) cos(x-y)= cos2x + cos2y - 1 ?

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To prove ( \cos(x+y) \cos(x-y) = \cos^2x + \cos^2y - 1 ), we'll use the trigonometric identity ( \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} ).

Starting with the left-hand side: [ \cos(x+y) \cos(x-y) ]

Using the sum and difference of angles formula for cosine: [ = (\cos x \cos y - \sin x \sin y)(\cos x \cos y + \sin x \sin y) ]

Expanding the expression: [ = \cos^2 x \cos^2 y - \sin^2 x \sin^2 y ]

Now, using the trigonometric identity ( \sin^2 \theta = 1 - \cos^2 \theta ): [ = \cos^2 x (1 - \cos^2 y) - (1 - \cos^2 x)\sin^2 y ]

Expanding further: [ = \cos^2 x - \cos^2 x \cos^2 y - \sin^2 y + \cos^2 x \sin^2 y ]

Using the identity ( \sin^2 \theta = 1 - \cos^2 \theta ) again: [ = \cos^2 x - \cos^2 x \cos^2 y - (1 - \cos^2 x) + \cos^2 x (1 - \cos^2 x) ]

Simplifying: [ = \cos^2 x - \cos^2 x \cos^2 y - 1 + \cos^2 x - \cos^4 x ]

Combining like terms: [ = 2\cos^2 x - \cos^2 x \cos^2 y - 1 - \cos^4 x ]

Using the identity ( \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} ) for ( \cos^2 x ): [ = 2\left(\frac{1 + \cos(2x)}{2}\right) - \cos^2 x \cos^2 y - 1 - \cos^4 x ]

[ = 1 + \cos(2x) - \cos^2 x \cos^2 y - 1 - \cos^4 x ]

[ = \cos(2x) - \cos^2 x \cos^2 y - \cos^4 x ]

[ = \cos^2 x + \cos^2 y - 1 ]

Therefore, ( \cos(x+y) \cos(x-y) = \cos^2 x + \cos^2 y - 1 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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