Question

How do you find the exact value $\sin \left( {x + y} \right)$ if $\cos x = \dfrac{8}{{17}}$ and $\sin y = \dfrac{{12}}{{37}}$?

Answer 1

Hint: Looking at the part to be found you will easily guess the formula required for it. But we do not have the entire values that are to be substituted in the formula. Hence you can use the trigonometric identities to get the remaining values using the values which are given to us already.

Complete step by step answer:
As given in the above question, we need to find $\sin \left( {x + y} \right)$.
We have a formula to find the value of $\sin \left( {x + y} \right)$ which goes as
$\sin \left( {x + y} \right) = \sin x\cos y + \cos x\sin y$
To get the value of $\sin \left( {x + y} \right)$ we just need to substitute the required values in the above formula.
But we have only two value i.e. $\cos x = \dfrac{8}{{17}}$ and $\sin y = \dfrac{{12}}{{37}}$. So first we will have to find the values of $\sin x$ and $\cos y$.
We can do this by using the trigonometric identity ${\sin ^2}x + {\cos ^2}x = 1$. Since we have the value of $\cos x = \dfrac{8}{{17}}$ we can get the value of $\sin x$
Hence substituting the value in the identity we get
$
  {\sin ^2}x + {\cos ^2}x = 1 \\
   \Rightarrow {\sin ^2}x = 1 - {\cos ^2}x \\
   \Rightarrow {\sin ^2}x = 1 - {\left( {\dfrac{8}{{17}}} \right)^2} \\
   \Rightarrow {\sin ^2}x = 1 - \dfrac{{64}}{{289}} \\
   \Rightarrow {\sin ^2}x = \dfrac{{289 - 64}}{{289}} \\
   \Rightarrow {\sin ^2}x = \dfrac{{225}}{{289}} \\
 $
Taking square root we get
\[\sin x = \dfrac{{15}}{{17}} - - - \left( 1 \right)\]
Also we need to find the value of $\cos y$. We can use the same identity ${\sin ^2}y + {\cos ^2}y = 1$.
So solving further we get,
  \[
  {\sin ^2}y + {\cos ^2}y = 1 \\
   \Rightarrow {\cos ^2}y = 1 - {\sin ^2}y \\
   \Rightarrow {\cos ^2}y = 1 - {\left( {\dfrac{{12}}{{37}}} \right)^2} \\
   \Rightarrow {\cos ^2}y = 1 - \dfrac{{144}}{{1369}} \\
   \Rightarrow {\cos ^2}y = \dfrac{{1369 - 144}}{{1369}} \\
   \Rightarrow {\cos ^2}y = \dfrac{{1225}}{{1369}} \\
 \]
Taking square root we get
$\cos y = \dfrac{{35}}{{37}} - - - \left( 2 \right)$
Hence we have obtained all the values.
$\sin \left( {x + y} \right) = \sin x\cos y + \cos x\sin y$
$
   = \dfrac{{15}}{{17}}.\dfrac{{35}}{{37}} + \dfrac{{12}}{{37}}.\dfrac{8}{{17}} \\
   = \dfrac{{621}}{{629}} \\
 $

hence the value of $\sin \left( {x + y} \right)$ is $\dfrac{{621}}{{629}}$ respectively.

Note: Similar to this question other questions regarding cos can also be asked. The only way to proceed the sum is to actually know the formula of the part to be found in this type of question. Following to which the trigonometric identities can be used as required in the formula from the values given to us.