prove 3cos theta - cos3theta = 2cos theta cos2theta + 4sin theta sin 2theta
To prove the equation 3cos(theta) - cos(3theta) = 2cos(theta)cos(2theta) + 4sin(theta)sin(2theta), we'll start by using trigonometric identities to simplify each side of the equation.
1. First, let's simplify the left side of the equation: 3cos(theta) - cos(3theta)
Using the triple-angle formula for cosine, cos(3theta) = 4cos^3(theta) - 3cos(theta), we can rewrite the left side:
3cos(theta) - cos(3theta) = 3cos(theta) - (4cos^3(theta) - 3cos(theta))
Simplifying further, we have:
3cos(theta) - cos(3theta) = 3cos(theta) - 4cos^3(theta) + 3cos(theta)
Combining like terms, we get:
3cos(theta) - cos(3theta) = 6cos(theta) - 4cos^3(theta)
2. Now, let's simplify the right side of the equation: 2cos(theta)cos(2theta) + 4sin(theta)sin(2theta)
Using the double-angle formulas for cosine and sine, cos(2theta) = 2cos^2(theta) - 1 and sin(2theta) = 2sin(theta)cos(theta), we can rewrite the right side:
2cos(theta)cos(2theta) + 4sin(theta)sin(2theta) = 2cos(theta)(2cos^2(theta) - 1) + 4sin(theta)(2sin(theta)cos(theta))
Simplifying further, we have:
2cos(theta)cos(2theta) + 4sin(theta)sin(2theta) = 4cos^3(theta) - 2cos(theta) + 8sin^2(theta)cos(theta)
Since sin^2(theta) = 1 - cos^2(theta), we can substitute this value and simplify:
2cos(theta)cos(2theta) + 4sin(theta)sin(2theta) = 4cos^3(theta) - 2cos(theta) + 8(1 - cos^2(theta))cos(theta)
Simplifying further, we have:
2cos(theta)cos(2theta) + 4sin(theta)sin(2theta) = 4cos^3(theta) - 2cos(theta) + 8cos(theta) - 8cos^3(theta)
Combining like terms, we get:
2cos(theta)cos(2theta) + 4sin(theta)sin(2theta) = 6cos(theta) - 4cos^3(theta)
3. Comparing the simplified expressions of the left and right sides, we find that they are equal:
3cos(theta) - cos(3theta) = 2cos(theta)cos(2theta) + 4sin(theta)sin(2theta)
Therefore, the equation 3cos(theta) - cos(3theta) = 2cos(theta)cos(2theta) + 4sin(theta)sin(2theta) is proven to be true.