Question

If \[\cos \theta =\dfrac{4}{5}\] and \[\cos \phi =\dfrac{12}{13}\], where \[\theta \] and \[\phi \] both lie in the fourth quadrant, find the values of \[\cos \left( \theta +\phi \right)\].

Answer 1

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

Complete step-by-step answer:

We are given the values of \[\cos \theta =\dfrac{4}{5}\] and \[\cos \phi =\dfrac{12}{13}\]. Now we will apply the formula of \[\cos \left( \theta +\phi \right)=\cos \theta \cos \phi -\sin \theta \sin \phi \].

Clearly, we can notice that we know the values of \[\cos \theta =\dfrac{4}{5}\] and \[\cos \phi =\dfrac{12}{13}\] and we are now to find the values of \[\sin \theta \] and \[\sin \phi \]. For this, we will 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 \[\cos \theta =\dfrac{4}{5}\] so by identity formula, we have,
\[si{{n}^{2}}\theta +{{\cos }^{2}}\theta =1....(1)\]

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

\[si{{n}^{2}}\theta +{{\left( \dfrac{4}{5} \right)}^{2}}=1\]

Now we will take the fraction \[{{\left( \dfrac{4}{5} \right)}^{2}}\] to the right side of the equality sign. Therefore, we have \[si{{n}^{2}}\theta =1-{{\left( \dfrac{4}{5} \right)}^{2}}\].

We know that \[{{\left( 4 \right)}^{2}}=16\] and \[{{\left( 5 \right)}^{2}}=25\]. Thus we have that,

\[si{{n}^{2}}\theta =1-\left( \dfrac{16}{25} \right)\].

Now we will take LCM on \[\left( \dfrac{16}{25} \right)\]. Since the LCM is clearly 25, therefore we have,

\[\begin{align}

  & {{\sin }^{2}}\theta =\dfrac{25-16}{25} \\

 & {{\sin }^{2}}\theta =\dfrac{9}{25} \\

\end{align}\]

or \[\sin \theta =\pm \sqrt{\dfrac{9}{25}}\].

As we know that the square root of 9 is 3 and the square root of 25 is 5. Thus we get,
\[\sin \theta =\pm \dfrac{3}{5}\].

According to the question, we have that \[\theta \] and \[\phi \] lies in the fourth quadrant and we know that the value of \[\sin \theta =\dfrac{-3}{5}\] instead of \[\dfrac{3}{5}\].

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

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

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

Therefore, we have,

\[\begin{align}

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

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

\end{align}\]

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

\[\begin{align}

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

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

 & {{\sin }^{2}}\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 \[\phi \] lies in the fourth quadrant and in the fourth quadrant, the value of sin is negative. 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{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}\]

Hence the value of \[\cos \left( \theta +\phi \right)\] is \[\dfrac{33}{65}\].

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 \[\sqrt{169}=13\].