Question

If $\cos \alpha +\cos \beta =a$, $\sin \alpha +\sin \beta =b$ and $\alpha -\beta =2\theta $ then $\dfrac{\cos 3\theta }{\cos \theta }=?$

Answer 1

Hint: We find the square values of the equations $\cos \alpha +\cos \beta =a$, $\sin \alpha +\sin \beta =b$. We add them to find the value of $2\cos 2\theta $. We simplify the equation $\dfrac{\cos 3\theta }{\cos \theta }$ to form it with $2\cos 2\theta $ and then place the value to get the final solution.

Complete step by step solution:
We assume $\cos \alpha +\cos \beta =a..........(i)$, $\sin \alpha +\sin \beta =b..........(ii)$.
We take squares of both equations and get
$\begin{align}
  & {{\left( \cos \alpha +\cos \beta \right)}^{2}}={{a}^{2}} \\
 & \Rightarrow {{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +2\cos \alpha \cos \beta ={{a}^{2}} \\
\end{align}$
$\begin{align}
  & {{\left( \sin \alpha +\sin \beta \right)}^{2}}={{b}^{2}} \\
 & \Rightarrow {{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +2\sin \alpha \sin \beta ={{b}^{2}} \\
\end{align}$
Now we add these two equations of squares to get
$\begin{align}
  & {{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +2\cos \alpha \cos \beta +{{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +2\sin \alpha \sin \beta ={{a}^{2}}+{{b}^{2}} \\
 & \Rightarrow 2+2\left( \cos \alpha \cos \beta +\sin \alpha \sin \beta \right)={{a}^{2}}+{{b}^{2}} \\
\end{align}$
We have the associative theorem of $\cos \alpha \cos \beta +\sin \alpha \sin \beta =\cos \left( \alpha -\beta \right)$.
We put the values and get $2+2\cos \left( \alpha -\beta \right)={{a}^{2}}+{{b}^{2}}$.
We also have that $\alpha -\beta =2\theta $ which gives
$\begin{align}
  & 2+2\cos \left( \alpha -\beta \right)={{a}^{2}}+{{b}^{2}} \\
 & \Rightarrow 2+2\cos 2\theta ={{a}^{2}}+{{b}^{2}} \\
 & \Rightarrow 2\cos 2\theta ={{a}^{2}}+{{b}^{2}}-2 \\
\end{align}$
We simplify the equation $\dfrac{\cos 3\theta }{\cos \theta }$ using the multiple angle formula of $\cos 3\theta =4{{\cos }^{3}}\theta -3\cos \theta $.
So, $\dfrac{\cos 3\theta }{\cos \theta }=\dfrac{4{{\cos }^{3}}\theta -3\cos \theta }{\cos \theta }=4{{\cos }^{2}}\theta -3$.
We also have $2{{\cos }^{2}}\theta =1+\cos 2\theta $.
Changing the equation, we get $\dfrac{\cos 3\theta }{\cos \theta }=4{{\cos }^{2}}\theta -3=2\left( 1+\cos 2\theta \right)-3=2\cos 2\theta -1$.
We now replace the value of $2\cos 2\theta ={{a}^{2}}+{{b}^{2}}-2$ in the equation of $\dfrac{\cos 3\theta }{\cos \theta }=2\cos 2\theta -1$.
So, $\dfrac{\cos 3\theta }{\cos \theta }=2\cos 2\theta -1={{a}^{2}}+{{b}^{2}}-2-1={{a}^{2}}+{{b}^{2}}-3$.
Therefore, the value of $\dfrac{\cos 3\theta }{\cos \theta }$ is ${{a}^{2}}+{{b}^{2}}-3$.

Note:
We need to remember that we can also divide with $\cos \theta $ assuming that $\cos \theta \ne 0$. We also need to remember that we used the identity formula of ${{\cos }^{2}}x+{{\sin }^{2}}x=1$ to find the simplified form.