Prove sin(4x)= (4sinxcosx)(2cos(x)^(2)-1
To prove the identity sin(4x) = (4sin(x)cos(x))(2cos(x)^2 - 1), we can start by using double-angle formula sin(2x) = 2sin(x)cos(x). Let's break it down step by step:
1. Start with the left-hand side (LHS) of the equation: sin(4x).
- This is the sine function applied to the angle 4x.
2. Apply the double-angle formula for sine on the LHS: sin(4x) = 2sin(2x)cos(2x).
- Using double-angle formula sin(2x) = 2sin(x)cos(x), we substitute 2x for x.
3. Simplify the expression on the RHS: sin(4x) = 2[2sin(x)cos(x)][cos^2(x) - sin^2(x)].
- By substituting sin(2x) = 2sin(x)cos(x) in step 2 and using the double-angle formula for cosine: cos(2x) = cos^2(x) - sin^2(x).
4. Continue simplifying: sin(4x) = 2[2sin(x)cos(x)](cos^2(x) - sin^2(x)).
- Combine the terms inside the square brackets.
5. Expand the expression further: sin(4x) = 4sin(x)cos(x)(cos^2(x) - sin^2(x)).
- Distribute the 2 to both terms inside the square brackets.
6. Simplify the expression inside the parentheses: sin(4x) = 4sin(x)cos(x)(2cos^2(x) - 1).
- Apply the distributive property.
Now we have shown that sin(4x) = 4sin(x)cos(x)(2cos^2(x) - 1), which proves the given identity.
Think of how the right hand side looks like the product of 2 sin(2x) and cos(2x)
The next step should be obvious: use the double angle identity again.