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

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! :)

Solution 1:

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

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

Solution 2:

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

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

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

Solution 3:

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

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

Solution 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

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