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))#
