Question

If $\theta $ lies in the first quadrant and $\cos \theta =\dfrac{8}{17}$, the prove that: $\cos \left( \dfrac{\pi }{6}+\theta \right)+\cos \left( \dfrac{\pi }{4}-\theta \right)+\cos \left( \dfrac{2\pi }{3}-\theta \right)=\left( \dfrac{\sqrt{3}-1}{2}+\dfrac{1}{\sqrt{2}} \right)\dfrac{23}{17}$

Answer 1

Hint: We will be using the concept of trigonometric identities to solve the problem. We will be using the trigonometric identity that $\cos \left( x+y \right)=\cos x\cos y-\sin x\sin y$. Then we will be using $\cos \theta =\dfrac{8}{17}$ to find the value of $\sin \theta $.

Complete step-by-step answer:

Now, we have been given that $\theta $ lies in the first quadrant and $\cos \theta =\dfrac{8}{17}$.

Now, we will take LHS and prove it to be equal to RHS. Now, in LHS we have,

$\cos \left( \dfrac{\pi }{6}+\theta \right)+\cos \left( \dfrac{\pi }{4}-\theta \right)+\cos \left( \dfrac{2\pi }{3}-\theta \right)$

Now, we will use the trigonometric identity that,

$\begin{align}

  & \cos \left( A+B \right)=\cos A\cos B-\sin A\sin B \\

 & \cos \left( A-B \right)=\cos A\cos B+\sin A\sin B \\

\end{align}$

So, we have,

$\cos \left( \dfrac{\pi }{6} \right)\cos \theta -\sin \left( \dfrac{\pi }{6} \right)\sin \theta +\cos \left( \dfrac{\pi }{4} \right)\cos \theta +\sin \left( \dfrac{\pi }{4} \right)\sin \theta +\cos \left( \dfrac{2\pi }{3} \right)\cos \theta +\sin \left( \dfrac{2\pi }{3} \right)\sin \theta $

Now, we know that the value of,

\[\begin{align}

  & \cos \left( \dfrac{\pi }{6} \right)=\dfrac{\sqrt{3}}{2} \\

 & \sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2} \\

 & \cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}} \\

 & \sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}} \\

 & \cos \left( \dfrac{2\pi }{3} \right)=\dfrac{-1}{2} \\

 & \sin \left( \dfrac{2\pi }{3} \right)=\dfrac{\sqrt{3}}{2} \\

\end{align}\]

So, using this we have,

\[=\dfrac{\sqrt{3}}{2}\cos \theta -\dfrac{1}{2}\sin \theta +\dfrac{1}{\sqrt{2}}\cos \theta +\dfrac{1}{\sqrt{2}}\sin \theta -\dfrac{1}{2}\cos \theta +\dfrac{\sqrt{3}}{2}\sin \theta \]

Now, we will collect the sine terms and cosine terms together. So, we have,

\[=\cos \theta \left( \dfrac{\sqrt{3}}{2}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{2} \right)+\sin \theta \left( \dfrac{\sqrt{3}}{2}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{2} \right)\]

Now, we will take the terms \[\left( \dfrac{\sqrt{3}-1}{2}+\dfrac{1}{\sqrt{2}} \right)\] common. So, we have,

\[\left( \dfrac{\sqrt{3}-1}{2}+\dfrac{1}{\sqrt{2}} \right)\left( \cos \theta +\sin \theta \right)\]
Now, we know that $\cos \theta =\dfrac{8}{17}$. So, we will use the trigonometric identity that ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ to find the value of $\sin \theta $. So, we have,

$\begin{align}

  & {{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 \\

 & {{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \\

 & \sin \theta =\pm \sqrt{1-{{\cos }^{2}}\theta } \\

\end{align}$

Now, we will use the value of $\cos \theta =\dfrac{8}{17}$ to find the value of $\sin \theta $.

$\begin{align}

  & \sin \theta =\pm \sqrt{1-\dfrac{64}{289}} \\

 & =\pm \sqrt{\dfrac{289-64}{289}} \\

 & =\pm \sqrt{\dfrac{225}{289}} \\

 & \sin \theta =\pm \dfrac{15}{17} \\

\end{align}$

Now, we have $\theta $ lies in the first quadrant. Therefore, $\sin \theta $ will be positive, so

we reject the value $\sin \theta =-\dfrac{15}{17}$.

Now, using $\sin \theta =\dfrac{15}{17}$ we have,

\[\begin{align}

  & \left( \dfrac{\sqrt{3}-1}{2}+\dfrac{1}{\sqrt{2}} \right)\left( \dfrac{15}{17}+\dfrac{8}{17} \right) \\

 & \left( \dfrac{\sqrt{3}-1}{2}+\dfrac{1}{\sqrt{2}} \right)\left( \dfrac{23}{17} \right) \\
\end{align}\]

Since, LHS = RHS.

Hence, proved.

Note: To solve these type of questions it is important to note that we have used the value of $\cos \theta =\dfrac{8}{17}$ and the formula ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ to find the value of $\cos \theta $. Also, we have used trigonometric formulae like,

$\begin{align}

  & \cos \left( A+B \right)=\cos A\cos B-\sin A\sin B \\

 & \cos \left( A-B \right)=\cos A\cos B+\sin A\sin B \\

\end{align}$