Establish the following identity
sin 2x/sinx - cos2x/cosx = secx
Use sin 2x = 2sinxcosx and
cos 2x = 2cos^2x - 1
in the left side, it comes apart nicely
To establish the given identity, we need to simplify the left-hand side (LHS) of the equation until it matches the right-hand side (RHS), which is sec(x).
Let's begin by simplifying each term in the LHS separately and then combining them.
First, let's simplify sin(2x)/sin(x):
We know the double-angle identity for sine, which states that sin(2x) = 2sin(x)cos(x).
So, we can rewrite sin(2x)/sin(x) as (2sin(x)cos(x))/sin(x).
Now, we can cancel out the sin(x) terms:
(2sin(x)cos(x))/sin(x) = 2cos(x).
Next, let's simplify cos(2x)/cos(x):
We know the double-angle identity for cosine, which states that cos(2x) = cos^2(x) - sin^2(x).
So, cos(2x)/cos(x) can be written as (cos^2(x) - sin^2(x))/cos(x).
Now, we can use the Pythagorean identity sin^2(x) + cos^2(x) = 1 to simplify further:
(cos^2(x) - sin^2(x))/cos(x) = (cos^2(x) - (1 - cos^2(x)))/cos(x) = (2cos^2(x) - 1)/cos(x).
Now, let's combine the two simplified terms:
LHS = 2cos(x) - (2cos^2(x) - 1)/cos(x).
To simplify this further, we need to find a common denominator for the terms. The common denominator is cos(x).
Taking the common denominator, we get:
LHS = (2cos(x) * cos(x) - (2cos^2(x) - 1))/cos(x)
Simplifying this:
LHS = (2cos^2(x) - 2cos^2(x) + 1)/cos(x) = (1)/cos(x) = sec(x).
Therefore, we have shown that the LHS is equal to the RHS, and the given identity sin(2x)/sin(x) - cos(2x)/cos(x) = sec(x) is established.