Question

# If $\sin \theta =\dfrac{3}{5}$ and $\cos \phi =\dfrac{-12}{13}$, where $\theta$ and $\phi$ both lies in the second quadrant. Find the values of $\cos \left( \theta +\phi \right)$.

Hint: We will apply the formula of trigonometry, $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ to simplify the question and we will apply the formula ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ to finally find the value of $\cos \left( \theta +\phi \right)$.

We are given the values of $\sin \theta =\dfrac{3}{5}$ and $\cos \phi =\dfrac{-12}{13}$. Now we will apply the formula of $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ to the term $\cos \left( \theta +\phi \right)$.

Therefore, we have $\cos \left( \theta +\phi \right)=\cos \theta \cos \phi -\sin \theta \sin \phi$.

Clearly, we can notice that we know the values of $\sin \theta =\dfrac{3}{5}$ and $\cos \phi =\dfrac{-12}{13}$ and we are now to find the values of $\sin \phi$ 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 $\sin \theta =\dfrac{3}{5}$ so by the identity formula we have,

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

We will substitute the value of $\sin \theta$ in equation (1). Thus we have ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.

$\Rightarrow {{\left( \dfrac{3}{5} \right)}^{2}}+{{\cos }^{2}}\theta =1$

Now we will take the fraction ${{\left( \dfrac{3}{5} \right)}^{2}}$to the right side of the equal sign.

Therefore, we have ${{\cos }^{2}}\theta =1-{{\left( \dfrac{3}{5} \right)}^{2}}$.

We know that ${{\left( 3 \right)}^{2}}=9$ and ${{\left( 5 \right)}^{2}}=25$. Thus, we have that,

${{\cos }^{2}}\theta =1-\dfrac{9}{25}$

Now we will take the LCM on $=1-\dfrac{9}{25}$. Since the LCM is clearly 25. Therefore, we have,

\begin{align} & {{\cos }^{2}}\theta =\dfrac{25-9}{25} \\ & {{\cos }^{2}}\theta =\dfrac{16}{25} \\ & \cos \theta =\pm \sqrt{\dfrac{16}{25}} \\ \end{align}

As we know that the square root of 16 is 4 and the square root of 25 is 5. Thus, we get,
$\cos \theta =\pm \dfrac{4}{5}$.

According to the question, we have that $\theta$ and $\phi$ lies in the second quadrant and we know that the value of $\cos \theta$ in the second quadrant is negative. So, the value of $\cos \theta =\dfrac{-4}{5}$ instead of $\dfrac{4}{5}$.

Similarly, we know that the value of $\cos \phi =\dfrac{-12}{13}$ and we need to find out the value of $\sin \phi$ = ?

This can be done using the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. Therefore, we have ${{\sin }^{2}}\phi +{{\cos }^{2}}\phi =1$.

Now we will substitute the value of $\cos \phi =\dfrac{-12}{13}$ and we get,

$si{{n}^{2}}\phi +{{\left( \dfrac{-12}{13} \right)}^{2}}=1$

Now we take the value ${{\left( \dfrac{-12}{13} \right)}^{2}}$ to the right side of the equality sign. Therefore, we have,

\begin{align} & \Rightarrow si{{n}^{2}}\phi =1-{{\left( \dfrac{-12}{13} \right)}^{2}} \\ & \therefore si{{n}^{2}}\phi =1-\dfrac{144}{169} \\ \end{align}

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

\begin{align} & si{{n}^{2}}\phi =\dfrac{169-144}{169} \\ & si{{n}^{2}}\phi =\dfrac{25}{169} \\ & si{{n}^{2}}\phi =\pm \sqrt{\dfrac{25}{169}} \\ \end{align}

Since, we know that $\sqrt{25}=5$ and $\sqrt{169}=13$. Thus we have,

$si{{n}^{2}}\phi =\pm \dfrac{5}{13}$

Since according to the question, we know that $\phi$ lies in the second quadrant and in the second quadrant, sin is positive. Therefore, we have,

$\sin \phi =\dfrac{5}{13}$

Hence the value of equation,

\begin{align} & \cos \left( \theta +\phi \right)=\cos \theta \cos \phi -\sin \theta \sin \phi \\ & \cos \left( \theta +\phi \right)=\left( \dfrac{-4}{5} \right)\left( \dfrac{-12}{13} \right)-\left( \dfrac{3}{5} \right)\left( \dfrac{5}{13} \right) \\ & \cos \left( \theta +\phi \right)=\dfrac{48}{65}-\dfrac{15}{65} \\ & \cos \left( \theta +\phi \right)=\dfrac{33}{65} \\ \end{align}

Note: We can apply HCF and LCM method for finding out the square root of the numbers. For example for finding the square root of 169 we will factorize it as $13\times 13$. This will also be the case in LCM and since 13 is a number that multiplies by itself to give 169. So we can write,

$\sqrt{169}=\sqrt{13\times 13}$ as $\sqrt{169}$ = 13.