Prove that for ans angle \( \theta \cdot \cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta \) (Hint: \( 3 \theta=\theta+2 \theta \) )

To prove the identity \( \cos 3\theta = 4 \cos^3 \theta - 3 \cos \theta \), we can use the hint provided: \( 3\theta = \theta + 2\theta \). Using the cosine angle addition formula, we have: \[ \cos(3\theta) = \cos(\theta + 2\theta) = \cos \theta \cos(2\theta) - \sin \theta \sin(2\theta) \] We now need to find \( \cos(2\theta) \) and \( \sin(2\theta) \): 1. From the double angle formulas: \[ \cos(2\theta) = 2 \cos^2 \theta - 1 \] \[ \sin(2\theta) = 2 \sin \theta \cos \theta \] 2. Substitute these into the expression for \( \cos(3\theta) \): \[ \cos(3\theta) = \cos \theta (2 \cos^2 \theta - 1) - \sin \theta (2 \sin \theta \cos \theta) \] \[ = \cos \theta (2 \cos^2 \theta - 1) - 2 \sin^2 \theta \cos \theta \] 3. Now, we know from the Pythagorean identity that \( \sin^2 \theta = 1 - \cos^2 \theta \). Substituting this into our equation gives: \[ = \cos \theta (2 \cos^2 \theta - 1) - 2 (1 - \cos^2 \theta) \cos \theta \] \[ = \cos \theta (2 \cos^2 \theta - 1) - 2 \cos \theta + 2 \cos^3 \theta \] \[ = 2 \cos^3 \theta - \cos \theta - 2 \cos \theta + 2 \cos^3 \theta \] \[ = 4 \cos^3 \theta - 3 \cos \theta \] This simplifies our identity: \[ \cos(3\theta) = 4 \cos^3 \theta - 3 \cos \theta \] So, the identity is proven!