Htykuut

I need some help with this exercise.

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

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! :)

$$frac{sin^{3}alpha - 2cos^{3}alpha + 3cosalpha}{3sinalpha +2cosalpha} = frac{sin^{3}alpha - 2cos^{3}alpha + 3cosalpha}{3sinalpha +2cosalpha} cdotfrac{1/cos^3alpha}{1/cos^3alpha} = frac{tan^3alpha-2+3cdot(1/cos^2alpha)}{(3tanalpha+2)cdot(1/cos^2alpha)}$$
Now, recall that $frac{1}{cos^2alpha}=sec^2alpha=1+tan^2alpha=5$, so,

$$frac{sin^{3}alpha - 2cos^{3}alpha + 3cosalpha}{3sinalpha +2cosalpha} = frac{tan^3alpha-2+3cdot(1/cos^2alpha)}{(3tanalpha+2)cdot(1/cos^2alpha)} = frac{8-2+15}{(6+2)5}=frac{21}{40}$$

Notice $boxed{sin alpha = 2cos alpha}$ and $$cos ^2alpha = {1over 1+tan^2alpha} ={1over 5}$$
so we have
$$frac{sin^{3}alpha - 2cos^{3}alpha + 3cosalpha}{3sinalpha +2cosalpha}=frac{8cos^{3}alpha - 2cos^{3}alpha + 3cosalpha}{6cosalpha +2cosalpha}$$

$$= frac{6cos^{3}alpha + 3cosalpha}{8cosalpha} = frac{6cos^{2}alpha + 3}{8} = {21over 40}$$

$$frac{sin^{3}alpha - 2cos^{3}alpha + 3cosalpha}{3sinalpha +2cosalpha}=frac{tan^{3}alpha - 2+ 3sec^2alpha}{(3tanalpha +2)sec^2alpha}$$

where $$sec^2alpha=tan^2alpha+1.$$

Hence $$frac{21}{40}.$$

Note that

$$tan alpha = frac{sin alpha}{cos alpha}$$

Thus if

$$tan alpha = 2$$

then

$$sin alpha = 2 cos alpha$$

Now just plug for sine

$$frac{sin^3 alpha - 2 cos^3 alpha + 3 cos alpha}{3 sin alpha + 2 cos alpha} = frac{8 cos^3 alpha - 2 cos^3 alpha + 3 cos alpha}{6 cos alpha + 2 cos alpha}$$

which then simplifies to

$$frac{6 cos^3 alpha + 3 cos alpha}{8 cos alpha} = frac{1}{8} [6 cos^2 alpha + 3]$$

Now note that

$$frac{1}{cos alpha} = sec alpha$$

and we have the trigonometric identity

$$1 + tan^2 alpha = sec^2 alpha$$

thus $$sec^2 alpha = 1 + (2)^2 = 1 + 4 = 5$$ and thus $cos^2 alpha = frac{1}{5}$. Thus we can plug that into the prior expression to get

$$frac{sin^3 alpha - 2 cos^3 alpha + 3 cos alpha}{3 sin alpha + 2 cos alpha} = frac{1}{8} [6 cos^2 alpha + 3] = frac{1}{8} [6 frac{1}{5} + 3] = frac{21}{40}$$.

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 $$frac{sin^3(a)-2cos^3(a)+3cos(a)}{3sin(a)+2cos(a)}$$ 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. $$frac{frac{8}{5sqrt5}-frac{2}{5sqrt5}+frac{3}{sqrt5}}{frac{6}{sqrt5}+frac{2}{sqrt5}}=frac{frac{8}{5}-frac{2}{5}+3}{6+2}=frac{21}{40}$$

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.

If $tanalpha=2$, then $$sinalpha=frac{2}{sqrt{1+2^2}}=frac{2}{sqrt{5}}$$ &$$cosalpha=frac{1}{sqrt{5}} ,$$ which you can see if you image a right-angled triangle of sides (other than the hypotenuse - I forget what those sides are called) of length 1 & 2, with the one of length 2 facing the angle $alpha$, & you can get the answer thence by manipulating surds & fractions.

And all the occurences of $sin$ & $cos$ are with odd exponent - what matters with that is that they are all of the same parity, so there won't even be a $sqrt{5}$ in the final answer.