Find the exact value of $25\sin(\alpha)\cos(\alpha)+\frac{75}{4}\sin(\alpha)$, with $\tan \alpha = \frac{1}{3}$

Also, $\alpha \in ]0,\frac{\pi}{2}[$

I tried:

$$\alpha = \arctan \frac{1}{3}$$

And so:

$$25\sin(\arctan \frac{1}{3})\cos(\arctan \frac{1}{3})+\frac{75}{4}\sin(\arctan \frac{1}{3})$$

That didn't look very helpful, so I tried:

$$\frac{\sin(\alpha)}{\cos(\alpha)}=\frac{1}{3} \Leftrightarrow \cos(\alpha) = 3\sin(\alpha)$$

And so $$25\sin(\alpha)3\sin(\alpha)+\frac{75}{4}\sin(\alpha) = \\75\sin(\alpha)+\frac{75}{4}\sin(\alpha) =\\ 75(\sin(\alpha)+\frac{\sin(\alpha)}{4}) =\\ 75(\frac{5\sin(\alpha)}{4}) =\\ \frac{375\sin(\alpha)}{4} = \\ \frac{375\sin(\arctan(\frac{1}{3}))}{4} = ???$$

My book says the solution is $\frac{15}{2}+\frac{15}{8}\sqrt{10}$.

How do I solve this?

  • $\begingroup$ In the last set of lines, it should be $75\sin^2\alpha + \frac{75}{4}\sin\alpha$. $\endgroup$ – zahbaz Apr 23 '17 at 2:46

We have $$\tan\alpha =\frac{1}{3} =\frac{\text{opposite side}}{\text{adjacent side}}$$

Then construct a right angled triangle, with hypotenuse equal to $\sqrt{1^2+3^2} =\sqrt{10}$

Then $$\cos\alpha = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{3}{\sqrt{10}}$$

and $$\sin\alpha =\frac{\text{opposite}}{\text{hypotenuse}} =\frac{1}{\sqrt{10}}$$

Then we can compute $$f(\alpha) = 25\sin(\alpha)\cos(\alpha)+\frac{75}{4}\sin(\alpha) = 25\cdot\frac{1}{\sqrt{10}}\cdot\frac{3}{\sqrt{10}}+\frac{75}{4}\cdot\frac{1}{\sqrt{10}}$$

$$f(\alpha) = \frac{75}{10}+\frac{75}{4\sqrt{10}} = \frac{15}{2}+\frac{75\sqrt{10}}{40} =\frac{15}{2}+\frac{15\sqrt{10}}{8}$$

$$f(\alpha) = \frac{60+15\sqrt{10}}{8}$$


$$75\sin^2\alpha + \frac{75}{4}\sin\alpha$$

Draw a right triangle with alpha in one corner.

\begin{align} \frac{75}{10} + \frac{75\sqrt{10}}{4\cdot10} \\ \\ \frac{15}{2} + \frac{15\sqrt{10}}{8} \end{align}

  • $\begingroup$ How did you get from $\frac{1}{\sqrt{10}}$ to $\frac{\sqrt{10}}{10}$? $\endgroup$ – Mark Read Apr 23 '17 at 3:04
  • 1
    $\begingroup$ Nevermind, I got it. $\endgroup$ – Mark Read Apr 23 '17 at 3:04

Hint -

Make a right angle triangle. We know that $\tan \theta = \frac{\text{perpendicular}} {\text{base}}$.

Comparing it with $\tan \alpha$ you get perpendicular = 1 and base = 3.

Now find 3rd side of triangle using phythagorous theorem. Then you can find value of $\cos \alpha$ and $\sin \alpha$.


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