hello

I am thinking this should factorise as the difference of 2 squares or be converted using trig identities

I just keep hitting dead ends though

http://img704.imageshack.us/img704/9982/sinms.jpg

Any help would be appreciated

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- Feb 15th 2010, 10:17 PM200001solving a trig equation
hello

I am thinking this should factorise as the difference of 2 squares or be converted using trig identities

I just keep hitting dead ends though

http://img704.imageshack.us/img704/9982/sinms.jpg

Any help would be appreciated - Feb 16th 2010, 12:12 AMOpalg
- Feb 16th 2010, 12:26 AM200001
Great, that sorts it

how do i decide on future problems what to divide by? - Feb 16th 2010, 05:13 AMSoroban
Hello, 200001!

This isa difference of two squares . . .*not*

Quote:

$\displaystyle 3\sin^2\!\theta - \sin\theta\cos\theta - 4\cos^2\!\theta \:=\:0$

Factor: .$\displaystyle (\sin\theta + \cos\theta)(3\sin\theta - 4\cos\theta) \:=\:0$

And we have two equations to solve:

$\displaystyle \sin\theta \;+\; \cos\theta \:=\:0 \quad\Rightarrow\quad \sin\theta \:=\:-\cos\theta \quad\Rightarrow\quad \frac{\sin\theta}{\cos\theta} \:=\:-1 \quad\Rightarrow\quad \tan\theta \:=\:-1$

. . Hence: .$\displaystyle \theta \:=\:\arctan(-1)$

$\displaystyle 3\sin\theta \;-\; 4\cos\theta \:=\:0 \quad\Rightarrow\quad 3\sin\theta \:=\:4\cos\theta \quad\Rightarrow\quad \frac{\sin\theta}{\cos\theta} \:=\:\frac{4}{3} \quad\Rightarrow\quad \tan\theta \:=\:\frac{4}{3}$

. . Hence: .$\displaystyle \theta \:=\:\arctan\left(\tfrac{4}{3}\right)$

- Feb 17th 2010, 05:29 PMDotdash13
With fairly basic trig, you pretty much have less than 10 options total, and you can usually rule several of those out just by looking at what trig functions are being used. for example, in your problem you could have also divided by $\displaystyle sin^2$, but you would not have really wanted to use something like secant to tangent. You learn mostly with practice.