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)(-33/65)# and #y=tan^(-1)(13/35)#, then we really have #sin(x+y)#. And if you recall, #sin(x+y)=sinxcosy+sinycosx#. So all we really need to do is find what #sinx#, #siny#, #cosx#, and #cosy# are equal to.

We already know what #cosx# is equal to. If #x=cos^(-1)(-33/65)#, then #cosx=-33/65#, using the inverse relationship between cosine and inverse cosine. But how do we find #sinx#? Simple - draw a triangle! Remember that #cosx="adjacent"/"hypotenuse"#, so #cosx=-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:

#"hypotenuse"^2="adjacent"^2+"opposite"^2#

#65^2=(-33)^2+"opposite"^2#

#3136="opposite"^2#

#56="opposite"#

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

From the diagram, it is clear that #sinx#, which is opposite divided by hypotenuse, is #56/65#. We have #sinx# and #cosx#, now we only need #siny# and #cosy#. Finding them will be almost the same thing.

If #y=tan^(-1)(13/35)#, then #tany=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:

#"hypotenuse"^2="adjacent"^2+"opposite"^2#

Solving for the hypotenuse, we get:

#"hypotenuse"^2=(35)^2+(13)^2#

#"hypotenuse"^2=1394#

#"hypotenuse"=sqrt(1394)#

Here's this triangle:

We can see that #siny=13/sqrt(1394)# and #cosy=35/sqrt(1394)#.

Let's remind ourselves of what we found so far:

#sinx=56/65#

#cosx=-33/65#

#siny=13/sqrt(1394)#

#cosy=35/sqrt(1394)#

Using the sum formula for sine, we can obtain our result:

#sin(x+y)=sinxcosy+sinycosx#

#->sin(cos^(-1)(-33/65)+tan^(-1)(13/35))=(56/65)(35/sqrt(1394))+(13/sqrt(1394))(-33/65)#

#->sin(cos^(-1)(-33/65)+tan^(-1)(13/35))=1960/(65sqrt(1394))-429/(65sqrt(1394))#

#->sin(cos^(-1)(-33/65)+tan^(-1)(13/35))=1960/(65sqrt(1394))-429/(65sqrt(1394))#

#->sin(cos^(-1)(-33/65)+tan^(-1)(13/35))=1531/(65sqrt(1394))#