sin(4x)/sin(x)=4cos(x)cos(x)
verify the identity
sin 4 x = 2 sin 2x cos 2x
= 2 (2 sin x cos x)(2cos^2 x - 1)
= 4 sin x (2cos^2 x - 1) cos x
now identity
4 (2cos^2 x - 1) cos x =?4cos^2x
(2cos^2 x -1) = cos x ??
Perhaps I am missing something?
It is important that the question be copied correctly, whether for posts or for exams.
The question should read:
sin(4x)/sin(x)=4cos(x)cos(2x)
Reduce each side to cos(x) and the identity will be obvious.
To verify the given trigonometric identity:
sin(4x)/sin(x) = 4cos(x)cos(x)
We will use trigonometric identities and algebraic manipulations to simplify both sides of the equation and show that they are equal.
First, let's work on the left-hand side (LHS):
sin(4x)/sin(x)
We can rewrite sin(4x) as sin(2x + 2x) using the double angle identity:
sin(2θ) = 2sin(θ)cos(θ)
Therefore,
sin(4x) = 2sin(2x)cos(2x)
Substituting this back into the equation:
2sin(2x)cos(2x) / sin(x)
Now, let's simplify the right-hand side (RHS):
4cos(x)cos(x) = 4cos^2(x)
Now, we have:
2sin(2x)cos(2x) / sin(x) = 4cos^2(x)
To prove the identity, we need to simplify both sides to the same expression. In this case, both sides are equal to 4cos^2(x).
Using trigonometric identities, we can simplify the LHS further:
2sin(2x)cos(2x) = sin(4x)
Therefore, we have:
sin(4x) / sin(x) = 4cos^2(x)
Both sides are now equal to 4cos^2(x), which verifies the given identity.
Note: It's important to fully understand the trigonometric identities and algebraic properties to solve and verify trigonometric identities effectively.