Question

Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation.
(sin x)(tan x cos x - cot x cos x) = 1 - 2 cos2x

Answer 1

Is that a square on the right side or a (2x) inside of cosine?

Answer 2

@zepdrix, probably a \(\cos^2x\). Expanding the left side gives a variant of \(\cos^2x-\sin^2x\).

Answer 3

@maywadee313,
\[\sin x\left(\tan x\cos x-\cot x\cos x\right)=\cdots\\
\sin x\left(\frac{\sin x}{\cos x}\cos x-\frac{\cos x}{\sin x}\cos x\right)=\cdots\\
\sin x\left(\sin x-\frac{\cos^2 x}{\sin x}\right)=\cdots\\
\sin^2 x-\cos^2 x=\cdots
\]
Use the double angle identity for cosine for the next few steps:
\[\cos2x=\cos^2x-\sin^2x\\~~~~~~~~~=2\cos^2x-1\]

Answer 4

@SithsAndGiggles , whats the double angle identity?

Answer 5

It's the last thing I listed: \(\cos2x=\cos^2x-\sin^2x\).

Answer 6

but how do I use it? I never learned that

Answer 7

Hey, you know what? Forget I mentioned "double angle" identity. You don't need to know about it for this specific problem. Just getting ahead of myself :) If you haven't learned it yet, you will eventually.

Do you understand how I got to \(\sin^2x-\cos^2x\) in the last step?

Answer 8

kind of. mind explaining?

Answer 9

From first to second line, I simply used the definitions of \(\tan x\) and \(\cot x\):
\[\tan x=\frac{\sin x}{\cos x}~~~\text{and}~~~\cot x=\frac{1}{\tan x}=\frac{\cos x}{\sin x}\]
From second to third, I simplified the quantity inside the parentheses.

From third to fourth, I distributed the sine on the outside.

Answer 10

alright I understand that

Answer 11

Okay, so the last few steps require using the Pythagorean identity: \(\sin^2x+\cos^2x=1\). You're familiar with that one, right?
\[\sin^2x-\cos^2x=\left(1-\cos^2x\right)-\cos^2x\]

Answer 12

yes. I am familiar with that for the most part.

Answer 13

Okay, so the last step is to just combine like terms. Left side matches up with right, so question answered:
\[1-\cos^2x-\cos^2x=1-2\cos^2x\]

Answer 14

ohhh okay :) thanks !

Answer 15

You're welcome