# More trigonometric equations

• Jun 8th 2008, 11:42 PM
littlelisa
More trigonometric equations
Once again, I need help on some trigonometric equations.

If cos(x)=-3/5, and pi/2 < x < 3pi/2, find sin(x/2)

Rewrite as product: sin3y-sin4y

If sin2x-cosx=0, find all the solutions in the interval [0, 2pi).

Thanks.

(If I don't respond it's because I've fallen asleep)
• Jun 8th 2008, 11:56 PM
Moo
Hello,

Quote:

Originally Posted by littlelisa
Once again, I need help on some trigonometric equations.

If cos(x)=-3/5, and pi/2 < x < 3pi/2, find sin(x/2)

You know this formula :

$\displaystyle \cos 2x=1-2 \sin^2 x$

Therefore, if we replace x by x/2, we get :

$\displaystyle \cos x=1-2 \sin^2 \frac x2$

$\displaystyle \implies \sin \frac x2=\pm \sqrt{\frac{1-\cos x}{2}}$

$\displaystyle \sin \frac x2=\pm \sqrt{\frac 45}$

Because pi/2 < x < 3pi/2, pi/4 < x/2 < 3pi/4, which is in the first or the second quadrant. So sin x/2 is positive.

$\displaystyle \boxed{\sin \frac x2=\sqrt{\frac 45}=\frac{2}{\sqrt{5}}}$
• Jun 9th 2008, 12:13 AM
Moo
Quote:

Originally Posted by littlelisa

Rewrite as product: sin3y-sin4y

hmm

Formulae you can learn and that will be useful =)

$\displaystyle \sin x+\sin y=2 \sin \frac{x+y}{2} \cdot \cos \frac{x-y}{2}$
$\displaystyle \sin x-\sin y=2 \cos \frac{x+y}{2} \cdot \sin \frac{x-y}{2}$

$\displaystyle \cos x+\cos y=2 \cos \frac{x+y}{2} \cdot \cos \frac{x-y}{2}$
$\displaystyle \cos x-\cos y=2 \sin \frac{x+y}{2} \cdot \sin \frac{x-y}{2}$

For the last 2 ones, you can remember that the cosine is always racist, so it will be a product of cosines only or sines only. It goes the same way for the addition formula, cosines remain with cosines and so do sines.

Here, just use the second formula :)
You can try to demonstrate them ^^

Quote:

If sin2x-cosx=0, find all the solutions in the interval [0, 2pi).

Thanks.

(If I don't respond it's because I've fallen asleep)
rewrite :

$\displaystyle \sin 2x=2 \cos x \sin x$

So you have to solve :

$\displaystyle 2 \cos x \sin x-\cos x=0$

$\displaystyle \implies \cos x \left(2 \sin x-1\right)=0$

Can you take it from here ?
• Jun 9th 2008, 02:36 PM
littlelisa
Thanks for the help. What you did in the last post kinda confused me a because I don't think I've learned those formulas, but I'm gonna try and figure them out. Thanks again!