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

Sin(a+b) Sin(a-b) Formula, Proof

FAQs

Q1: What is the identity of cos^3x-sin^3x?

Answer: The identity of cos3x -sin3x is given by cos3x -sin3x = (cosx -sinx)(2+sin2x)/2.

Q2: What is the identity of cos^3x + sin^3x?

Answer: The identity of cos3x +sin3x is given by cos3x +sin3x = (cosx +sinx)(2-sin2x)/2.

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