# How solve a trigometric function

• May 9th 2009, 10:09 PM
Stealth
How solve a trigometric function
Hey everybody.

I have the following trig function and I have to simplify it so that it will only have sintheta in it. Can you show me how to do that?:

9.78049(1+0.005288sinsquaredx - 0.000006sinsquared2x) (m/ssquared)

I tried pulling out a sinx, but that only seemed to complicate the problem because once I did that had to simplify the sin2x and I just couldn't find a way to do it!

Is there another way to approach this problem? I've exhausted all sources and my book has the most basic concepts possible!

Any help is appreciated,
Stealth
• May 10th 2009, 04:10 AM
skeeter
Quote:

Originally Posted by Stealth
Hey everybody.

I have the following trig function and I have to simplify it so that it will only have sintheta in it. Can you show me how to do that?:

9.78049(1+0.005288sinsquaredx - 0.000006sinsquared2x) (m/ssquared)

theta?

$\displaystyle a[1 + b\sin^2{x} - c\sin^2(2x)]$

$\displaystyle a[1 + b\sin^2{x} - 4c\sin^2{x}\cos^2{x}]$

$\displaystyle a[1 + b\sin^2{x} - 4c\sin^2{x}(1 - \sin^2{x})]$

$\displaystyle a[1 + b\sin^2{x} - 4c\sin^2{x} + 4c\sin^4{x}]$

$\displaystyle a[1 + (b-4c)\sin^2{x} + 4c\sin^4{x}]$
• May 10th 2009, 07:00 AM
Stealth
Quote:

Originally Posted by skeeter
theta?

$\displaystyle a[1 + b\sin^2{x} - c\sin^2(2x)]$

$\displaystyle a[1 + b\sin^2{x} - 4c\sin^2{x}\cos^2{x}]$

$\displaystyle a[1 + b\sin^2{x} - 4c\sin^2{x}(1 - \sin^2{x})]$

$\displaystyle a[1 + b\sin^2{x} - 4c\sin^2{x} + 4c\sin^4{x}]$

$\displaystyle a[1 + (b-4c)\sin^2{x} + 4c\sin^4{x}]$

Sorry yes it's x not theta. I replaced it with x to make it easier on my eyes. Um....so I can't simplify it anymore right? I can't drop it down to sinx only?
• May 10th 2009, 09:21 AM
skeeter
Quote:

Originally Posted by Stealth
Sorry yes it's x not theta. I replaced it with x to make it easier on my eyes. Um....so I can't simplify it anymore right? I can't drop it down to sinx only?

no
• May 10th 2009, 09:57 AM
Stealth
Ok so I'm not the only one that can't solve it anymore...
Thanks! (Happy)

Stealth