1. ## 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)

2. Hello,

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.

$\displaystyle \boxed{\sin \frac x2=\sqrt{\frac 45}=\frac{2}{\sqrt{5}}}$

3. 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 ^^

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 ?

4. 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!