Prove (sin4x)/(1-cos4x) = cotx - 1/2(cscxsecx)
Please help me get started on this question... I'm at a lost as for what to do first....
sin4x/(1-cos4x)
(2 sin2x cos2x)/(2sin^2 2x)
cos2x/sin2x
(2cos^2 x - 1)/(2sinx cosx)
cosx/sinx - 1/(2sinx cosx)
cotx - 1/2 cscx secx
To prove the given trigonometric equality, we will start by simplifying the left-hand side (LHS) and the right-hand side (RHS) separately.
Let's start with the LHS:
(sin(4x))/(1 - cos(4x))
We can use the double-angle formula for sine and the double-angle formula for cosine to write the LHS in terms of x:
LHS = (2sin(2x)cos(2x))/(1 - (2cos^2(2x) - 1))
= (2sin(2x)cos(2x))/(-2cos^2(2x))
= -sin(2x)/(cos(2x))
Now, let's simplify the RHS:
cot(x) - (1/2)(csc(x)sec(x))
Using the reciprocal trigonometric identities, we can rewrite cot(x), csc(x), and sec(x) in terms of sine and cosine:
cot(x) = cos(x)/sin(x)
csc(x) = 1/sin(x)
sec(x) = 1/cos(x)
Substitute these identities into the RHS:
RHS = (cos(x)/sin(x)) - (1/2)((1/sin(x))(1/cos(x)))
= (cos(x)/sin(x)) - (1/2)(1/(sin(x)cos(x)))
= (cos(x)/sin(x)) - (1/(2sin(x)cos(x)))
= (cos(x)/sin(x)) - (1/sin(2x))
= (cos(x)/sin(x)) - csc(2x)
Now, let's focus on the LHS and RHS and try to bring them to a common form.
LHS = -sin(2x)/(cos(2x))
RHS = (cos(x)/sin(x)) - csc(2x)
To bring them to a common form, we will use the double-angle formula for sine and cosine:
sin(2x) = 2sin(x)cos(x)
cos(2x) = cos^2(x) - sin^2(x)
Substitute these identities into the LHS:
LHS = -(2sin(x)cos(x))/(cos^2(x) - sin^2(x))
Now, let's simplify the RHS to match this form:
RHS = (cos(x)/sin(x)) - (1/sin(2x))
= (cos(x)/sin(x)) - (1/(2sin(x)cos(x)))
= (2cos^2(x)/(2sin(x)cos(x))) - (1/(2sin(x)cos(x)))
= (2cos^2(x) - 1)/(2sin(x)cos(x))
Finally, we compare the simplified LHS and RHS:
LHS = -(2sin(x)cos(x))/(cos^2(x) - sin^2(x))
RHS = (2cos^2(x) - 1)/(2sin(x)cos(x))
They are now in a common form, and we can see that LHS = RHS. Therefore, we have proved the given trigonometric equality:
(sin(4x))/(1 - cos(4x)) = cot(x) - (1/2)(csc(x)sec(x))