# Thread: Some help with proving trig identities and other things

1. ## Some help with proving trig identities and other things

I need some help with proving some trig identities step-by-step, justifying each step with either the word 'algebra' or the name of the fundamental trig identity used, and changing only one side to match the other (no adding one to both sides, for example)
The first is
(1 + csc x)/ sec x = cos x + cot x,
which I can see will involve cos x = 1/sec x somehow, but I can't see how to make cot x into 1 + csc x.
The next is
tan(x - π/4) = tan x - 1/1 + tan x,
which I can see may involve the cofunction identities, though again I don't see how.

The last is
cos x/sec x - 1 - cos x/ sec x + 1 = 2cot^2 x * cos x

I've simplified the left side to 2cosx/sec^2x - 1, which may or may not be helpful.

The next set of problems I'm having difficulty with is finding all solutions to both of the following equations the equations on the interval [0, 2pi).
I simply have no idea how to do these- what's the domain/range (whichever's applicable) for these functions again? Does it even matter?
2 cos x + sin 2x = 0

cot(x/2) = 1 - cos x/1 + cos x

Help me out!
Toptomcat

2. Originally Posted by Toptomcat
The first is
(1 + csc x)/ sec x = cos x + cot x,
You have,
$\displaystyle \frac{1+\csc x}{\sec x}=\cos x+\cot x$
On the Left hand side express reciptocals,
$\displaystyle \frac{1+\frac{1}{\sin x}}{\frac{1}{\cos x}}$
$\displaystyle \frac{\frac{1+\sin x}{\sin x}}{\frac{1}{\cos x}}$
Divide fractions (remember to flip),
$\displaystyle \frac{1+\sin x}{\sin x}\cdot \frac{\cos x}{1}$
Mutiply to get,
$\displaystyle \frac{\cos x+\sin x\cos x}{\sin x}$
Express as two seperate fractions,
$\displaystyle \frac{\cos x}{\sin x}+\frac{\sin x\cos x}{\sin x}$
Cancel and simplyfy,
$\displaystyle \cot x+\cos x$
Q.E.D.

3. Originally Posted by Toptomcat
The next is
tan(x - π/4) = tan x - 1/1 + tan x,
which I can see may involve the cofunction identities, though again I don't see how.
Use the tangent sum-difference rule,
$\displaystyle \tan(x\pm y)=\frac{\tan x\pm \tan y}{1\mp \tan x\tan y}$
Thus,
$\displaystyle \tan (x-\pi/4)=\frac{\tan x- \tan (\pi/4)}{1+ \tan x\tan (\pi/4)}=\frac{\tan x-1}{1+\tan x}$
But you lacked parantheses you show have written,
Code:
(tanx-1)/(tanx+1)