Question

If \[\theta \] and \[\phi \] lie in the first quadrant, such that \[\sin \theta =\dfrac{8}{17}\] and \[\cos \phi =\dfrac{12}{13}\], find the values of \[\cos \left( \theta +\phi \right)\].

Answer 1

Hint: We will apply the formula of trigonometry \[\cos \left( A+B \right)=\operatorname{cosAcosB}-sinAsinB\] to simplify the question and we will apply the formula \[si{{n}^{2}}\theta +{{\cos }^{2}}\theta =1\] to finally find the value of \[\cos \left( \theta +\phi \right)\].

Complete step-by-step answer:

We are given the values of \[\sin \theta =\dfrac{8}{17}\] and \[\cos \phi =\dfrac{12}{13}\]. Now we will apply the formula of \[\cos \left( A+B \right)=\operatorname{cosAcosB}-sinAsinB\].

Clearly, we can notice that the know the values of \[\cos \phi =\dfrac{12}{13}\] and \[\sin \theta =\dfrac{8}{17}\] and we are now to find the values of \[\sin \phi \] and \[\cos \theta \]. For this, we apply a trigonometric identity. Here, we use trigonometric identity given as \[si{{n}^{2}}\theta +{{\cos }^{2}}\theta =1\]. As we know the value of \[\sin \theta =\dfrac{8}{17}\], so by identity formula, we have,

\[si{{n}^{2}}\theta +{{\cos }^{2}}\theta =1.....(1)\]

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

\[\Rightarrow {{\left( \dfrac{8}{17} \right)}^{2}}+{{\cos }^{2}}\theta =1\]

Now we will take the fraction \[{{\left( \dfrac{8}{17} \right)}^{2}}\] to the right hand side of the equal sign. Therefore, we have \[{{\cos }^{2}}\theta =1-{{\left( \dfrac{8}{17} \right)}^{2}}\]. We know that \[{{\left( 8 \right)}^{2}}=64\] and \[{{\left( 17 \right)}^{2}}=289\]. Thus we have that \[{{\cos }^{2}}\theta =1-\dfrac{64}{289}\].

Now we will take LCM on \[1-\dfrac{64}{289}\]. Since the LCM is clearly 289. Therefore, we have,

\[\begin{align}

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

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

 & \cos \theta =\sqrt{\dfrac{225}{289}} \\

\end{align}\]

As we know that the square root of 225 is 15 and the square root of 289 is 17. Thus we get, \[\cos \theta =\pm \dfrac{15}{17}\].

According to the question, we have that \[\theta \] and \[\phi \] lies in the first quadrant and we know that the value of \[\cos \theta \] is positive in the first quadrant.

So the value of \[\cos \theta =\dfrac{15}{17}\] instead of \[-\dfrac{15}{17}\].

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 trigonometric identity \[si{{n}^{2}}\theta +{{\cos }^{2}}\theta =1\]. Therefore, we have the substitution of the value of \[\cos \phi =\dfrac{12}{13}\] and we get,

\[\sin \phi +{{\left( \dfrac{12}{13} \right)}^{2}}=1\].

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

\[\begin{align}

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

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

\end{align}\]

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

\[\begin{align}

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

 & {{\sin }^{2}}\phi =\dfrac{25}{169} \\

 & \sin \phi =\pm \sqrt{\dfrac{25}{169}} \\

\end{align}\]

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

\[\sin \phi =\pm \dfrac{5}{13}\].

Since according to the question, we know that \[\theta \] lies in the first quadrant and in the first quadrant, the value of sin is positive. Therefore, we have,

\[\sin \phi =\dfrac{5}{13}\].


Hence, the value of the equation,

\[\begin{align}

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

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

 & \cos \left( \theta +\phi \right)=\dfrac{180}{221}-\dfrac{40}{221} \\

 & \cos \left( \theta +\phi \right)=\dfrac{140}{221} \\

\end{align}\]

Hence the value of \[\cos \left( \theta +\phi \right)=\dfrac{140}{221}\].

Note: We can apply HCF or LCM method for finding the square root of the numbers. For example, for finding the square root of 169, we will factor 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 numbers. So we can write, \[\sqrt{169}=\sqrt{13\times 13}\] as \[\sqrt{169}=13\]. Also, here it is important to read the question carefully. As it is given that the angles lie in the first quadrant, we must choose only positive values as all the angles are positive in the first quadrant.