(cos 2x)/(1/(cos x)) * (sin(pi + x))/(tan x) = (sec(x) - csc(x)) * (csc(x))/(sec^2 x) * (1 - cos^2 x)
My friend and I are having a debate on the true identity of this equation, however my friend learned this subject and I did not. Could someone maybe shed some light on how to find the identity and the steps ?
You are talking about Julia's post here:
https://www.jiskha.com/questions/1829582/prove-the-identity-of-the-following-equation-cos-2x-1-cos-x-sin-pi-x-tan-x
Take a look what I did, and note my comments.
I think there is a typo
To solve this equation and prove its identity, we will start by simplifying both sides step by step. Let's begin:
Starting with the left-hand side (LHS):
1. Use the identity cos(pi + x) = -cos(x):
LHS = (cos 2x) / (1 / cos x) * (sin(pi + x)) / (tan x)
LHS = (cos 2x) / (1 / cos x) * (-sin x) / (tan x)
LHS = (cos 2x) * (cos x) * (-sin x) / (1 * sin x)
LHS = -cos 2x * cos x
Continuing with the right-hand side (RHS):
2. Use the identity sec(x) = 1 / cos(x), csc(x) = 1 / sin(x), and cos^2(x) = 1 - sin^2(x):
RHS = (sec(x) - csc(x)) * (csc(x)) / (sec^2(x)) * (1 - cos^2(x))
RHS = (1 / cos(x) - 1 / sin(x)) * (1 / sin(x)) / (1 / cos^2(x)) * (1 - (1 - sin^2(x)))
RHS = (1 - sin(x) / sin(x)cos(x)) * (cos^2(x) / 1) * sin^2(x)
RHS = (1 - 1 / (cos(x) / (sin(x)))) * (cos^2(x)) * sin^2(x)
RHS = (1 - 1 / tan(x)) * (cos^2(x)) * sin^2(x)
RHS = (cos^2(x) / tan(x)) * (1 - tan^2(x))
RHS = cos^2(x) - (cos^2(x) \cdot tan^2(x))
Now, comparing the simplified left-hand side (LHS) and right-hand side (RHS), we can see that they are equal:
LHS = -cos 2x * cos x
RHS = cos^2(x) - (cos^2(x) \cdot tan^2(x))
Therefore, we have proven the identity:
-cos 2x * cos x = cos^2(x) - (cos^2(x) \cdot tan^2(x))