Question

If \[A+B+C=\pi \] then the value of \[\cos \left( A+B \right)+\cos C\] is equal to:
(a) 0
(b) \[2\cos C\]
(c) \[\cos C\sin C\]
(d) \[2\sin C\]

Answer 1

Hint: We solve this problem by using the half-angle formula.
We have the half angle formula of sum of trigonometric ratio that is
\[\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)\]
By using the above formula and the given condition that \[A+B+C=\pi \] we find the value of required expression.

Complete step by step answer:
We are given with a condition that is
\[A+B+C=\pi \]
We are asked to find the value of \[\cos \left( A+B \right)+\cos C\]
Let us assume that the given expression as
\[\Rightarrow A=\cos \left( A+B \right)+\cos C\]
Now, let us use the formula of half-angle of the sum of trigonometric ratio.
We know that the half angle formula of sum of trigonometric ratio that is
\[\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)\]
Now, by using the above formula to given expression we get
\[\Rightarrow A=2\cos \left( \dfrac{A+B+C}{2} \right)\cos \left( \dfrac{A+B-C}{2} \right)\]
Now, by substituting the value that is \[A+B+C=\pi \] in above equation we get
\[\Rightarrow A=2\cos \left( \dfrac{\pi }{2} \right)\cos \left( \dfrac{A+B-C}{2} \right)\]
We know that the standard value of the cosine ratio that is
\[\cos \dfrac{\pi }{2}=0\]
By substituting this value in above equation we get
\[\begin{align}
  & \Rightarrow A=2\times 0\times \cos \left( \dfrac{A+B-C}{2} \right) \\
 & \Rightarrow A=0 \\
\end{align}\]
Therefore we can conclude that the value of given expression is 0 that is
\[\therefore \cos \left( A+B \right)+\cos C=0\]
So, option (a) is correct answer.

Note:
We have other method to solve this problem
We are given with a condition that is
\[A+B+C=\pi \]
Now, let us take the value of \[A+B\] in terms of \[C\] then we get
\[\Rightarrow A+B=\pi -C\]
We are asked to find the value of \[\cos \left( A+B \right)+\cos C\]
Let us assume that the given expression as
\[\Rightarrow A=\cos \left( A+B \right)+\cos C\]
Now, by substituting the value of \[A+B\] in terms of \[C\] in above equation we get
\[\Rightarrow A=\cos \left( \pi -C \right)+\cos C\]
We know that the standard formula of cosine ratio that is
\[\cos \left( \pi -\theta \right)=-\cos \theta \]
By using this formula in above equation we get
\[\begin{align}
  & \Rightarrow A=-\cos C+\cos C \\
 & \Rightarrow A=0 \\
\end{align}\]
Therefore we can conclude that the value of given expression is 0 that is
\[\therefore \cos \left( A+B \right)+\cos C=0\]
So, option (a) is correct answer.