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# Prove cos (3x) = 4cos^{3} (x) -3cos (x)

Trigonometry helps in finding the measure of unknown dimensions of a right-angled triangle using formulas and identities based on this relationship.

## Answer: Hence proved that cos (3x) = 4cos^{3}x−3cosx

We have to prove cos (3x) = 4cos^{3}x−3cosx

**Expanantion :**

We know that,

**cos(A+B) = cos(A)cos(B) - sin(A)sin(B)**

We can write cos(3x) = cos(2x + x)

Using the formula let's solve LHS = cos(3x)

cos(2x + x) = cos2x cosx − sin2x sinx

= (−1+2cos^{2}x) cosx − (2cosx sinx) sinx (Since, cos2x = -1 + 2cos^{2}x and sin2x = 2cosx sinx)

= − cosx + 2 cos^{3}x − 2sin^{2}x cosx

= − cosx + 2cos^{3}x − 2(1 − cos^{2}x)cosx (Since, sin^{2}x = 1 - cos^{2}x)

= − cosx + 2 cos^{3}x - 2cosx + 2 cos^{3}x

= 4cos^{3}x − 3cosx = RHS

### Thus, verified cos (3x) = 4cos^{3}x − 3cosx

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