Some help with proving trig identities and other things

[Toptomcat]

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, 2π).
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

 

[pka]

\(\displaystyle \L
\frac{{1 + \csc (x)}}{{\sec (x)}} = \frac{1}{{\sec (x)}} + \frac{{\csc (x)}}{{\sec (x)}} = \cos (x) + \cot (x)\)

 

[Toptomcat]

Thanks a lot, that one's much clearer. Didn't think to apply algebra that way.
And the next?

 

[skeeter]

2cos(x) + sin(2x) = 0

use the double angle identity for sine ...

2cos(x) + 2sin(x)cos(x) = 0

factor ...

2cos(x)[1 + sin(x)] = 0

can you solve for x from here?


I'm assuming the last equation is ...
cot(x/2) = (1 - cos x)/(1 + cos x)

let u = x/2

cot(u) = [1 - cos(2u)]/[1 + cos(2u)]

use a couple of the double identities for cosine ...

cot(u) = [1 - (1 - 2sin²(u))]/[1 + (2cos²(u) - 1)]

cot(u) = [2sin²(u)]/[2cos²(u)]

cot(u) = tan²(u)

multiply both sides by tan(u) ...

1 = tan³(u)

0 = tan³(u) - 1

factor the cubic ...

0 = [tan(u) - 1][tan²(u) + tan(u) + 1]

the quadratic factor has a discriminant < 0 ... so it will not yield any real solutions

for the linear factor ...

tan(u) - 1 = 0

tan(u) = 1

since 0 < x < 2pi, 0 < x/2 < pi ... 0 < u < pi

tan(x/2) = 1 at x/2 = pi/4, so x = pi/2.

 

[soroban]

Hello, Toptomcat!

Prove: \(\displaystyle \L\,\tan(x\,-\,\frac{\pi}{4})\;=\;\frac{\tan x\,-\,1}{1\,+\,\tan x}\)

Use the formula: \(\displaystyle \,\tan(A\,-\,B)\;=\;\L\frac{\tan A\,-\,\tan B}{1\,+\,\tan A\cdot\tan B}\)

So we have: \(\displaystyle \:\tan\left(x\,-\,\frac{\pi}{4}\right) \;= \;\L\frac{\tan x\,-\,\tan\left(\frac{\pi}{4}\right)}{1\,+\,\tan x\cdot\tan\left(\frac{\pi}{4}\right)}\)

Since $\displaystyle \tan\left(\frac{\pi}{4}\right)\,-\,1$, we have: \(\displaystyle \L\,\frac{\tan x\,-\,1}{1\,+\,\tan x}\)

Prove: \(\displaystyle \L\,\frac{\cos x}{\sec x - 1}\,-\,\frac{\cos x}{\sec x\,+\,1}\;=\;2\cot^2 x\cdot\cos x\)

What you was good . . .

The left side is: \(\displaystyle \L\,\frac{\cos x(\sec x\,+\,1)\,-\,\cos x(\sec x\,-\,1)}{(\sec x\,-\,1)(\sec x\,+\,1)} \;=\;\frac{2\cdot\cos x}{\sec^2x\,-\,1} \;= \;\frac{2\cdot\cos x}{\tan^2x} \;=\;\frac{2\cdot\cos x}{\frac{\sin^2x}{\cos^2x}}\)

\(\displaystyle \L\;\;\;\;= \;\frac{2\cdot\cos^3x}{\sin^2x} \;= \;2\cdot\frac{\cos^2x}{\sin^2x}\cdot\frac{\cos x}{1}\;=\;2\left(\frac{\cos x}{\sin x}\right)^2\cdot\cos x \;= \;2\cdot\cot^2x\cdot\cos x\)

Solve for $\displaystyle x$ on $\displaystyle [0,\,2\pi):\;\;2\cdot\cos x\,+\,\sin 2x \;= \;0$

When a problem has two different angles (in this case $\displaystyle x$ and $\displaystyle 2x$),
$\displaystyle \;\;$use a Double-angle or Half-angle identity to make them the same.

For this one, we will use: \(\displaystyle \,\sin\2\theta \:=\:2\cdot\sin\theta\cdot\cos\theta\)

So we have: $\displaystyle \,2\cdot\cos x\,+\,2\cdot\sin x\cdot\cos x\:=\:0$

Factor: $\displaystyle \,2\cdot\cos x(1\,+\,\sin x)\:=\:0$

And we have two equations to solve;
$\displaystyle \;\;2\cdot\cos x\:=\:0\;\;\Rightarrow\;\;\cos x\:=\:0\;\;\Rightarrow\;\;x\,=\,\frac{\pi}{2},\;\frac{3\pi}{2}$
$\displaystyle \;\;1\,+\,\sin x\:=\:0\;\;\Rightarrow\;\;\sin x\:=\:-1\;\;\Rightarrow\;\;x\,=\,\frac{3\pi}{2}$

Solve for $\displaystyle x$ on $\displaystyle [0,\,2\pi):\;\;\cot\left(\frac{x}{2}\right)\;=\;\frac{1\,-\,\cos x}{1\,+\,\cos x}$

This one is messy . . .

We need: $\displaystyle \,\sin^2\left(\frac{x}{2}\right)\:=\:\frac{1\,-\,\cos x}{2}\;\;\Rightarrow\;\;1\,-\,\cos x\:=\:2\sin^2\left(\frac{x}{2}\right)$

$\displaystyle \;\;\;$ and: $\displaystyle \,\cos^2\left(\frac{x}{2}\right) \:=\:\frac{1\,+\,\cos x}{2}\;\;\Rightarrow\;\;1\,+\,\cos x\:=\:2\cos^2\left(\frac{x}{2}\right)$


The right side is: \(\displaystyle \L\:\frac{1\,-\,\cos x}{1\,+\,\cos x}\:=\:\frac{2\sin^2\left(\frac{x}{2}\right)}{2\cos^2\left(\frac{x}{2}\right)} \:=\:\left[\frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)\right]^2\)\(\displaystyle \:=\:2\tan^2\left(\frac{x}{2}\right)\)


The equation becomes: \(\displaystyle \L\:\cot\left(\frac{x}{2}\right)\:=\:\tan^2\left(\frac{x}{2}\right)\;\;\Rightarrow\;\;\frac{1}{\tan\left(\frac{x}{2}\right)}\:=\:\tan^2\left(\frac{x}{2}\right)\)

Then: \(\displaystyle \L\:\tan^3\left(\frac{x}{2}\right)\:=\:1\;\;\Rightarrow\;\;\tan\left(\frac{x}{2}\right)\:=\:1\;\;\Rightarrow\;\;\frac{x}{2}\:=\:45^o,\;225^o\)

$\displaystyle \;\;$Hence: \(\displaystyle \L\;x\:=\:90^o,\;450^o\)

But $\displaystyle 450^o$ is equivalent to $\displaystyle 90^o$.

$\displaystyle \;\;$Therefore, the only solution is: \(\displaystyle \L\,x\:=\:90^o\)