Given $\tan\alpha=2$, evaluate $\frac{\sin^{3}\alpha - 2\cos^{3}\alpha + 3\cos\alpha}{3\sin\alpha +2\cos\alpha}$

Question

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! :)

Answer 1

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

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

Answer 2

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

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

Answer 3

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

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

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

Answer 4

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

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

Answer 5

If $\tan\alpha=2$, then $$\sin\alpha=\frac{2}{\sqrt{1+2^2}}=\frac{2}{\sqrt{5}}$$ &$$\cos\alpha=\frac{1}{\sqrt{5}} ,$$ 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.

Answer 6

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.