# Question #86f6d

Jul 12, 2016

Draw a couple of triangles and make use of the sum formula for sine to get $\sin \left({\cos}^{- 1} \left(- \frac{33}{65}\right) + {\tan}^{- 1} \left(\frac{13}{35}\right)\right) = \frac{1531}{65 \sqrt{1394}}$.

#### Explanation:

These questions require us to remember what sine and cosine and tangent really are: relationships among legs of a right triangle. So, how would you find the sine of two different angles added together ? Draw two different right triangles using the information given in the problem and use the sum formula for sine.

If we let $x = {\cos}^{- 1} \left(- \frac{33}{65}\right)$ and $y = {\tan}^{- 1} \left(\frac{13}{35}\right)$, then we really have $\sin \left(x + y\right)$. And if you recall, $\sin \left(x + y\right) = \sin x \cos y + \sin y \cos x$. So all we really need to do is find what $\sin x$, $\sin y$, $\cos x$, and $\cos y$ are equal to.

We already know what $\cos x$ is equal to. If $x = {\cos}^{- 1} \left(- \frac{33}{65}\right)$, then $\cos x = - \frac{33}{65}$, using the inverse relationship between cosine and inverse cosine. But how do we find $\sin x$? Simple - draw a triangle! Remember that $\cos x = \text{adjacent"/"hypotenuse}$, so $\cos x = - \frac{33}{65}$ means we have a triangle with an adjacent side of $- 33$ and a hypotenuse of $65$. Using the Pythagorean Theorem, we can solve for the length of the other side:
${\text{hypotenuse"^2="adjacent"^2+"opposite}}^{2}$
${65}^{2} = {\left(- 33\right)}^{2} + {\text{opposite}}^{2}$
$3136 = {\text{opposite}}^{2}$
$56 = \text{opposite}$

Now that we know all the side lengths, we can construct the triangle:

From the diagram, it is clear that $\sin x$, which is opposite divided by hypotenuse, is $\frac{56}{65}$. We have $\sin x$ and $\cos x$, now we only need $\sin y$ and $\cos y$. Finding them will be almost the same thing.

If $y = {\tan}^{- 1} \left(\frac{13}{35}\right)$, then $\tan y = \frac{13}{35}$. Since tangent is defined as opposite over adjacent, we have a triangle with an opposite side of $13$ and an adjacent side of $35$. The Pythagorean Theorem says:
${\text{hypotenuse"^2="adjacent"^2+"opposite}}^{2}$

Solving for the hypotenuse, we get:
${\text{hypotenuse}}^{2} = {\left(35\right)}^{2} + {\left(13\right)}^{2}$
${\text{hypotenuse}}^{2} = 1394$
$\text{hypotenuse} = \sqrt{1394}$

Here's this triangle:

We can see that $\sin y = \frac{13}{\sqrt{1394}}$ and $\cos y = \frac{35}{\sqrt{1394}}$.

Let's remind ourselves of what we found so far:
$\sin x = \frac{56}{65}$
$\cos x = - \frac{33}{65}$
$\sin y = \frac{13}{\sqrt{1394}}$
$\cos y = \frac{35}{\sqrt{1394}}$

Using the sum formula for sine, we can obtain our result:
$\sin \left(x + y\right) = \sin x \cos y + \sin y \cos x$
$\to \sin \left({\cos}^{- 1} \left(- \frac{33}{65}\right) + {\tan}^{- 1} \left(\frac{13}{35}\right)\right) = \left(\frac{56}{65}\right) \left(\frac{35}{\sqrt{1394}}\right) + \left(\frac{13}{\sqrt{1394}}\right) \left(- \frac{33}{65}\right)$
$\to \sin \left({\cos}^{- 1} \left(- \frac{33}{65}\right) + {\tan}^{- 1} \left(\frac{13}{35}\right)\right) = \frac{1960}{65 \sqrt{1394}} - \frac{429}{65 \sqrt{1394}}$
$\to \sin \left({\cos}^{- 1} \left(- \frac{33}{65}\right) + {\tan}^{- 1} \left(\frac{13}{35}\right)\right) = \frac{1960}{65 \sqrt{1394}} - \frac{429}{65 \sqrt{1394}}$
$\to \sin \left({\cos}^{- 1} \left(- \frac{33}{65}\right) + {\tan}^{- 1} \left(\frac{13}{35}\right)\right) = \frac{1531}{65 \sqrt{1394}}$