Simplifying trigonometric expressions

**KillamSin** · January 29th 2009, 03:25 PM

I am having difficulties simplifying trigonometric expressions. My textbook isn't much help and I don't have a teacher to ask, so I thought maybe someone here could help me.

Some examples of questions I'm not really sure about are:

1) sinx / tanxcosx

The answer I got for this one simply "1".

2) sin(^4)x + (sin(^2)x)(cos(^2))

This one I got "sinx(^4)x + 1"

3)
sinx - (sinx)(cos(^2)x)
------------------------
sin(^2)x

I am lost on this one.

Any help would be greatly appreciated.

**anywho** · January 29th 2009, 04:19 PM

Here is number three...

$\frac{sin(x)-sin(x)*cos^2x}{sin^2x}$

First factor a sin(x) out of the numerator...

$\frac{sin(x)*(1-cos^2x)}{sin^2x}$

Then using the Pythagorean theorem we can say 1-cos^2x=sin^2x...

$\frac{sin(x)*sin^2(x)}{sin^2x}$

from there you can simplify to sin(x)

**Soroban** · January 29th 2009, 04:20 PM

Hello, KillamSin!

$1)\;\;\frac{\sin x}{\tan x\cos x}$

The answer I got for this one simply "1".

You are right!

$\frac{\sin x}{\tan x\cos x} \;=\;\frac{\sin x}{\dfrac{\sin x}{{\color{red}\rlap{/////}}\cos x}({\color{red}\rlap{/////}}\cos x)} \;=\;\frac{{\color{green}\rlap{////}}\sin x}{{\color{green}\rlap{////}}\sin x} \;=\;1$

$2)\;\; \sin^4\!x + \sin^2\!x\cos^2\!x$

$\text{Factor: }\;\sin^2\!x\underbrace{\left(\sin^2\!x + \cos^2\!x\right)}_{\text{This is 1}} \;=\;\sin^2\!x$

$3)\;\;\frac{\sin x - \sin x\cos^2\!x}{\sin^2\!x}$

$\text{Factor: }\;\frac{\sin x\overbrace{\left(1 - \cos^2\!x\right)}^{\text{This is }\sin^2\!x}}{\sin^2\!x} \;=\;\frac{\sin x\cdot{\color{red}\rlap{/////}}\sin^2\!x}{{\color{red}\rlap{/////}}\sin^2\!x} \;=\;\sin x$

**anywho** · January 29th 2009, 04:27 PM

that is some perty LaTex! me+LaTex=

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