how do you get sin(4theta)/1+cos(4theta) to equal tan(2theta)
To show that sin(4θ)/(1+cos(4θ)) is equal to tan(2θ), we'll use trigonometric identities to simplify and manipulate both expressions:
1. Rewrite the denominator:
sin(4θ) / (1 + cos(4θ)) = sin(4θ) / (cos(0) + cos(2 × 2θ))
= sin(4θ) / [2cos²(2θ)]
2. Let's represent tan(2θ) using the double-angle formula:
tan(2θ) = sin(2θ) / cos(2θ)
3. Apply the double-angle formulas for sin(2θ) and cos(2θ):
sin(2θ) = 2sin(θ)cos(θ)
cos(2θ) = cos²(θ) - sin²(θ)
4. Substitute these values into tan(2θ):
tan(2θ) = (2sin(θ)cos(θ)) / (cos²(θ) - sin²(θ))
= (2sin(θ)cos(θ)) / cos²(θ) - sin²(θ)
5. Rearrange and simplify:
tan(2θ) = 2(sin(θ) / cos(θ)) / (cos²(θ) / cos²(θ) - sin²(θ) / cos²(θ))
= 2tan(θ) / (1 - tan²(θ))
6. Now, let's substitute back θ = 2θ and rewrite the expression:
tan(2θ) = 2tan(θ) / (1 - tan²(θ))
tan(2θ) = 2tan(θ) / (1 - tan²(θ)) = sin(4θ) / [2cos²(2θ)]
Therefore, sin(4θ)/(1+cos(4θ)) is indeed equal to tan(2θ).