TITLE: Given $\tan\alpha=2$, evaluate $\frac{\sin^{3}\alpha - 2\cos^{3}\alpha + 3\cos\alpha}{3\sin\alpha +2\cos\alpha}$
QUESTION [10 upvotes]: I need some help with this exercise. 

Given that $$\tan\alpha=2$$
  calculate the value of:
  $$\frac{\sin^{3}\alpha - 2\cos^{3}\alpha + 3\cos\alpha}{3\sin\alpha +2\cos\alpha}$$

I've tried everything but I always end up having something irreducible in the end. Maybe this is an easy question, but I'm new at trigonometry. If anyone can help me by providing a step-by-step solution to this, I would be really thankful! :)

REPLY [2 votes]: I know that there are several good answers already, but I would like to show a helpful method that works in general.
Start with the equation $\tan(a) = 2$. It may be rearranged (see end) to $a = \arctan(2)$. The big fraction is $$\frac{\sin^3(a)-2\cos^3(a)+3\cos(a)}{3\sin(a)+2\cos(a)}$$ If the angle $a$ is substituted into all the sines and cosines, there will be two common expresions, $\sin(\arctan(2))$ and $\cos(\arctan(2))$
Imagine a triangle in which you find it’s angle to be $\arctan(2)$. That means that the opposite is twice as long as the adjacent. For our purposes, let the triangle have a horizontal adjacent leg of 1, and a vertical opposite leg of 2. In order to find the sine or cosine of this, the hypotenuse must be known. By the pythagorean theorem, the hypotenuse is $\sqrt5$. The sine of the angle of this triangle is the opposite, 2, divided by the hypotenuse, $\sqrt5$. $$\sin(\arctan(2)) = \frac{2}{\sqrt5}$$
The cosine of the angle in this triangle is the adjacent, 1, divided by the hypotenuse. $$\cos(\arctan(2)) = \frac{1}{\sqrt5}$$
All that’s left is to substitute these in and simplify. $$\frac{\frac{8}{5\sqrt5}-\frac{2}{5\sqrt5}+\frac{3}{\sqrt5}}{\frac{6}{\sqrt5}+\frac{2}{\sqrt5}}=\frac{\frac{8}{5}-\frac{2}{5}+3}{6+2}=\frac{21}{40}$$
Be careful with arctangent. Inverse trig functions have infinite values besides the one a calculator gives. The angles that satisfy $\tan(a)=2$ are spaced by 180° turns. But, the equivalent angles of sine and cosine are spaced by 360°. When 180° is added to the angle in sine and cosine, they give the exact negative. I made sure that the negatives could cancel in your fraction before doing this; it is all odd powers.