Question

If $\sin \varphi =\dfrac{15}{17}$ and $\cos \varphi =\dfrac{12}{13}$ , where both lie in the first quadrant. Find the values of $\cos \left( \theta -\varphi \right)$.

Answer 1

Hint: We will apply the formula of trigonometry $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$ and we will apply the formula.

Complete step-by-step answer:

We are given the values of $\sin \theta =\dfrac{15}{17}$and $\cos \varphi =\dfrac{12}{13}$. Now, we will apply the formula of $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$. Therefore, we have

$\begin{align}

  & \cos \left( \theta +\left( -\varphi \right) \right)=\cos \left( \theta -\varphi \right) \\

 & \Rightarrow \cos \left( \theta -\varphi \right)=\cos \theta \cos \varphi +\sin \theta \sin \varphi \\

\end{align}$

Now, since we know the value of $\sin \theta =\dfrac{15}{17}$ and $\cos \varphi =\dfrac{12}{13}$ .We will find the values of $\sin \varphi $ and $\cos \theta $. For this we will apply a trigonometric identity. Here, we use the trigonometric identity given as ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. As we know the value of $\sin \theta =\dfrac{15}{17}$ is by the identity formula we have

${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$……….(i)

We will substitute the value of $\sin \theta $ in (i). Thus, we have

$\begin{align}

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

 & \Rightarrow {{\left( \dfrac{15}{17} \right)}^{2}}+{{\cos }^{2}}\theta =1 \\

\end{align}$

Now we will take the fraction ${{\left( \dfrac{15}{17} \right)}^{2}}$ to the right side of the equal sign. Therefore, we have ${{\cos }^{2}}\theta =1-{{\left( \dfrac{15}{17} \right)}^{2}}$ . We know that ${{\left( 15 \right)}^{2}}=225$ and ${{\left( 17 \right)}^{2}}=289$ . Thus we have that

${{\cos }^{2}}\theta =1-\dfrac{225}{289}$

Now we will take the LCM on $1-\dfrac{225}{289}$ .Since, the LCM is clearly 289. Therefore, we have,

$\begin{align}

  & {{\cos }^{2}}\theta =\dfrac{289-225}{289} \\

 & \Rightarrow {{\cos }^{2}}\theta =\dfrac{64}{289} \\

 & \Rightarrow \cos \theta =\pm \sqrt{\dfrac{64}{289}} \\

\end{align}$

Since, we know that and $\sqrt{64}=8$ and square root 289 is 17. Thus, we get

$\cos \theta =\pm \dfrac{8}{17}$

According to the question, we have that $\theta $ and $\varphi $ lies in the first quadrant. And we know that the value of $\cos \theta $ in the first quadrant is positive. Therefore, $\cos \theta =\pm \dfrac{8}{17}$

Similarly, we know the value of $\cos \varphi =\dfrac{12}{13}$ and we need to find out the value of $\sin \varphi $ . This can be done using the trigonometric identity. Therefore, we have

${{\sin }^{2}}\varphi +{{\cos }^{2}}\varphi =1$

Now, we will substitute the value of $\cos \varphi =\dfrac{12}{13}$ and we get ${{\sin }^{2}}\varphi +{{\left( \dfrac{12}{13} \right)}^{2}}=1$

Now, we will take the value${{\left( \dfrac{12}{13} \right)}^{2}}$ to the right side of the equal sign. Therefore, we have\[\begin{align}

  & {{\sin }^{2}}\varphi =1-{{\left( \dfrac{12}{13} \right)}^{2}} \\

 & \Rightarrow {{\sin }^{2}}\varphi =1-\dfrac{144}{169} \\

\end{align}\]

This is because we know that square of 12=144 and square of 13=169. Now, we take LCM here.

$\begin{align}

  & \Rightarrow {{\sin }^{2}}\varphi =\dfrac{169-144}{169} \\

 & \Rightarrow {{\sin }^{2}}\varphi =\dfrac{25}{169} \\

 & \Rightarrow \sin \varphi =\pm \sqrt{\dfrac{25}{169}} \\

\end{align}$

We have, $\sin \varphi =\pm \dfrac{5}{13}$

Since, according to the question we know that the $\varphi $ lies in first quadrant and in first quadrant the value of $\sin \varphi $ is given by $\sin \varphi =\pm \dfrac{5}{13}$ .

Hence, the value of the equation,

\[\begin{align}

  & \cos \left( \theta +\left( -\varphi \right) \right)=\cos \theta \cos \varphi +\sin \theta \sin \varphi \\

 & \Rightarrow \cos \left( \theta -\varphi \right)=\left( \dfrac{8}{17} \right)\left( \dfrac{12}{13} \right)+\left( \dfrac{15}{17} \right)\left( \dfrac{5}{13} \right) \\

 & \Rightarrow \cos \left( \theta -\varphi \right)=\dfrac{96}{221}+\dfrac{75}{221} \\

 & \Rightarrow \cos \left( \theta -\varphi \right)=\dfrac{171}{221} \\

\end{align}\]

Hence, the value of \[\cos \left( \theta -\varphi \right)=\dfrac{171}{221}\] .

Note: We can apply the HCF or LCM method for finding out the square root of the numbers. For example: For finding the square root of 25 we will factor it as $5\times 5$. This will also be in case in LCM. And, since 5 is a number that is multiplied by itself only so it will be like $\sqrt{5\times 5}=5$.