Help! Proving trigonometic identities!

**StonerPenguin** · February 20th 2010, 05:46 PM

Ah, I've always sucked at doing proofs, so I'd like help with these two;
Prove that $sin^2 x (1 + cot^2x) = 1$
Prove that $tan x (cot x + tan x) = sec^2x$

Sorry that I haven't shown my beginning steps, I'll work on it while I wait for a response.

Thanks in advance for any help

**pickslides** · February 20th 2010, 06:11 PM

Originally Posted by StonerPenguin

Ah, I've always sucked at doing proofs, so I'd like help with these two;
Prove that $sin^2 x (1 + cot^2x) = 1$

$\sin^2 x (1 + \cot^2x)$

$=\sin^2 x +\sin^2 x \cot^2x$

$=\sin^2 x +\frac{\sin^2 x }{\tan^2x}$

$=\sin^2 x +\cos^2x$

$<br /> =1<br />$

**Gusbob** · February 20th 2010, 06:12 PM

Originally Posted by StonerPenguin

Prove that $sin^2 x (1 + cot^2x) = 1$

The easiest way is to use the identity $1 \cot^2x = cosec^2x$
If you're not allowed to assume that
$LHS = \sin^2 x (1 + cot^2x)$
$= \sin^2x \left(1 + \frac{\cos^2x}{\sin^2x} \right)$
$= \sin^2x \left(\frac{\sin^2x + \cos^2x}{\sin^2x}\right)$
Use the identity $\sin^2x + \cos^2x = 1$ and cancel out the $\sin^2x$ and you'll get the RHS.

Prove that $tan x (cot x + tan x) = sec^2x$

$LHS = tan x (cot x + tan x)$

$= \frac{\sin x}{\cos x} \left(\frac{\cos x}{\sin x} + \frac{\sin x}{\cos x} \right)$

$= \frac{\sin x}{\cos x} \left(\frac{\cos^2 x + \sin^2 x}{\sin x \cos x} \right)$

Again just use $\sin^2 x + \cos^2 x = 1$ and multiply out to get the RHS.

**StonerPenguin** · February 22nd 2010, 08:15 AM

Wow! Thanks guys! Lots of smart people from Austrailia!

I really appreciate the help, but Gusbob, the second one doesn't look right to me;

Originally Posted by Gusbob

$LHS = tan x (cot x + tan x)$

$= \frac{\sin x}{\cos x} \left(\frac{\cos x}{\sin x} + \frac{\sin x}{\cos x} \right)$

$= \frac{\sin x}{\cos x} \left(\frac{\cos^2 x + \sin^2 x}{\sin x \cos x} \right)$

Again just use $\sin^2 x + \cos^2 x = 1$ and multiply out to get the RHS.

I don't see how $\frac{\sin x}{\cos x} \left(\frac{\cos^2 x + \sin^2 x}{\sin x \cos x} \right) = \sin^2 x + \cos^2 x$

Doesn't $= \frac{\sin x}{\cos x} \left(\frac{\cos^2 x + \sin^2 x}{\sin x \cos x} \right) = \frac{1}{\cos x} \left(\frac{\cos^2 x + \sin^2 x}{\cos x} \right) = \frac{\cos^2 x + \sin^2 x}{\cos^2 x}$ ? And even if I'm being stupid and you are right, I'm suppose to transform the it into $sec^2x$ not 1. (I'm so sorry if I sound too harsh or ungrateful, I just want makes sure it's right D: )

Anywho, here's what I got, does this look right?

$\tan x (\cot x + \tan x) = \sec^2x$ Given
$\sec^2x = \tan x (\cot x + \tan x)$ Swapping the sides around
$= (\tan x \cot x)+ (\tan^2 x)$ Simplify by multiplying
$(\tan x \cot x) = 1$ Reciprocal Identity
$= 1 + \tan^2 x = \sec^2x$ Pythagorean Identity
$\sec^2x = \sec^2x$ TAH-DAH~

Also, here's two other question I'd like to double check;
Find the exact value of cos 105° without using tables or a calculator.
(Using the addition formula for cosine)
$\cos 105° = \cos (60° + 45°)$
$= \cos 60° \cos 45° - \sin 60° \sin 45°$
$= (\frac{1}{2} \left)(\frac{\sqrt2}{2} \right) - (\frac{\sqrt3}{2} \left)(\frac{\sqrt2}{2} \right)$
$= \frac{\sqrt2}{4} - \frac{\sqrt6}{4}= \frac{\sqrt2 - \sqrt6}{4}$

Is that right?

And, "Find the exact value of tan 15° without using tables or a calculator."

(Using a Half-Angle formula for the tangent)
$\tan 15° = \tan \frac{30}{2}$
$= \frac{1 - \cos 30}{\sin 30}$
${1} - \frac{\sqrt3}{2}$
_______
.... $\frac{1}{2}$
= ${2} - {\sqrt3}$

BTW How do you put subordinate fractions using the math bbcode? That was the only way I could think to do it.

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