sin^4 x = (3-4cos 2x + cos4x)/8
prove on one side please!
RS
= (3 - 4cos(2x) + cos^2(2x) - sin^2(2x))/8
= (3 - 4cos(2x) + cos^2(2x) - (1 - cos^2(2x)))/8
= (2cos^2(2x) - 4cos(2x) + 2)/8
= (2(cos^2(2x) - 2cos(2x) + 1))/8
= (2(cos(2x) - 1)^2)/8
= (2(1 - 2sin^2(x) - 1)^2)/8
= (2(-2sin^2(x))^2)/8
= (8sin^4(x))/8
= sin^4(x)
= LS
hope I made no typos this time
how?
1-4cos2x+cos4x/8=sin4x
To prove that sin^4(x) = (3 - 4cos(2x) + cos(4x))/8, we will simplify both sides of the equation separately and show that they are equal.
Starting with the left-hand side (LHS):
sin^4(x)
Now, let's simplify the right-hand side (RHS) of the equation:
(3 - 4cos(2x) + cos(4x))/8
To simplify further, we need to express cos(2x) and cos(4x) in terms of sin(x). We can use the identities:
cos(2x) = 1 - 2sin^2(x)
cos(4x) = 1 - 8sin^2(x) + 8sin^4(x)
Now, let's substitute these values into the RHS:
(3 - 4(1 - 2sin^2(x)) + (1 - 8sin^2(x) + 8sin^4(x)) / 8
Simplifying further, we have:
(3 - 4 + 8sin^2(x) + 1 - 8sin^2(x) + 8sin^4(x)) / 8
(-1 + 8sin^2(x) + 8sin^4(x)) / 8
(8sin^4(x) + 8sin^2(x) - 1) / 8
Now, we can see that the RHS of the equation is in the same form as the LHS of the equation (sin^4(x)). Therefore, we have shown that sin^4(x) = (3 - 4cos(2x) + cos(4x))/8, as required.