Simplify cos^3x-sin^3x | Formula of cos^3x+sin^3x

The simplification of cos³x -sin³x is equal to cos³x -sin³x = (cosx -sinx)(2+sin2x)/2. The formula of cos^3x +sin^3x is given as follows:

cos³x +sin³x = $\dfrac{(\cos x +\sin x)(2-\sin 2x)}{2}$.

Formula of cos³x -sin³x

Prove that cos³x -sin³x = (cosx -sinx)(2+sin2x)/2.

Solution:

Applying the formula a³ -b³ = (a -b)(a²+ab+b²), we get that

cos³x -sin³x = (cosx -sinx) (cos²x +cosx sinx + sin²x)

= (cosx -sinx) (cos²x + sin²x +cosx sinx)

= (cosx -sinx) (1+ $\dfrac{2\cos x \sin x}{2}$) by the identity cos²x + sin²x =1.

= (cosx -sinx) (1+ $\dfrac{\sin 2x}{2}$) by the formula sin2x = 2sinx cosx.

= $\dfrac{(\cos x -\sin x)(2+\sin 2x)}{2}$.

So the formula of cos³x -sin³x is given by cos³x -sin³x = (cosx -sinx)(2+sin2x)/2.

Read Also: Formula of cos⁴x +sin⁴x

Simplify cos³x +sin³x

Prove that cos³x +sin³x = (cosx +sinx)(2-sin2x)/2.

Solution:

By using the formula a³ +b³ = (a +b)(a² -ab+b²), it follows that

cos³x +sin³x = (cosx +sinx) (cos²x -cosx sinx + sin²x)

= (cosx +sinx) (cos²x + sin²x -cosx sinx)

= (cosx +sinx) (1- $\dfrac{2\cos x \sin x}{2}$)

= (cosx +sinx) (1- $\dfrac{\sin 2x}{2}$)

= $\dfrac{(\cos x +\sin x)(2-\sin 2x)}{2}$.

So the formula of cos³x +sin³x is given by cos³x +sin³x = (cosx +sinx)(2-sin2x)/2.

You Can Read: Sin3x Formula | Cos3x Formula

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