1. ## Trigonometry

Greetings,

I have the problem below. I've tried dividing everything to sina or cosa - I get one or several tga's, but eventually I end up with a single sina or cosa with some other numbers and from there I loop...

Thank you.

2. Hello,

If tga means tan(a), use the formula $\displaystyle 1+\tan^2 a=\frac{1}{\cos^2 a}$ and $\displaystyle \cos^2 a+\sin^2 a=1$.

Bye

3. I do not understand. I am left with an extra cosa again.

Indeed, tg is tan. I am thought to abreviate it as tg.

4. Hello,

Since tan a=-2 and $\displaystyle 1+\tan^2 a=\frac{1}{\cos^2 a}$,$\displaystyle \cos a=\pm\frac{1}{\sqrt{5}}$. Use $\displaystyle \tan a=\frac{\sin a}{\cos a}$ to get $\displaystyle \sin a=\mp\frac{1}{\sqrt{5}}$. Plug in everything.

Bye.

5. Originally Posted by Logic
Greetings,

I have the problem below. I've tried dividing everything to sina or cosa - I get one or several tga's, but eventually I end up with a single sina or cosa with some other numbers and from there I loop...

Thank you.
If the trigonometric functions don't want to cooperate, bypass them and use the reference triangle of the angle instead.

tan(a) = -2
That means "a" is in the 2nd or 4th quadrants.

opposite side = 2
hypotenuse = sqrt(5)
sin(a) = 2/sqrt(5)
cos(a) = -1/sqrt(5)
So,
sin(a) / [sin^3(a) +3cos^2(a)
= [2/sqrt(5)] / [(2/sqrt(5))^3 +3(-1/sqrt(5))^2]