If the given values are $A+B=\dfrac{\pi }{3}$ and $\cos A+\cos B=1$, then which following options are true:
(a) $\cos \left( A-B \right)=\dfrac{1}{3}$.
(b) $\left| \cos A-\cos B \right|=\sqrt{\dfrac{2}{3}}$.
(c) $\cos \left( A-B \right)=\dfrac{-1}{3}$.
(d) $\left| \cos A-\cos B \right|=\dfrac{1}{2\sqrt{3}}$.
Answer
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Hint: We use trigonometric transformation for $\cos A+\cos B=1$. After transforming the equation we substitute the given value $A+B=\dfrac{\pi }{3}$ in it and we do some suitable modifications to cosine function to find the value of $\cos \left( A-B \right)$. Now, to find the value of $\left| \cos A-\cos B \right|$, we use trigonometric transformation for equation present inside the modulus. We calculate all the required values by using trigonometric identities to get the result.
Complete step-by-step solution:
We have given that the values are $A+B=\dfrac{\pi }{3}$ and $\cos A+\cos B=1$. We need to check which are the true among the given options.
Let us keep $A+B=\dfrac{\pi }{3}$ as equation (1).
So, we have $\cos A+\cos B=1$.
From trigonometric transformations we know that $\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right).\cos \left( \dfrac{A-B}{2} \right)$.
We have $2\cos \left( \dfrac{A+B}{2} \right).\cos \left( \dfrac{A-B}{2} \right)=1$.
From the equation we have a value $A+B=\dfrac{\pi }{3}$, we substitute this in the above equation.
We have $2\cos \left( \dfrac{\dfrac{\pi }{3}}{2} \right).\cos \left( \dfrac{A-B}{2} \right)=1$.
We have $2\cos \left( \dfrac{\pi }{6} \right).\cos \left( \dfrac{A-B}{2} \right)=1$.
We know the value of $\cos \left( \dfrac{\pi }{6} \right)=\dfrac{\sqrt{3}}{2}$.
We have $2.\left( \dfrac{\sqrt{3}}{2} \right).\cos \left( \dfrac{A-B}{2} \right)=1$.
We have $\sqrt{3}.\cos \left( \dfrac{A-B}{2} \right)=1$.
We have $\cos \left( \dfrac{A-B}{2} \right)=\dfrac{1}{\sqrt{3}}$ -------(2).
Let us square equation (2) on both sides.
We have ${{\cos }^{2}}\left( \dfrac{A-B}{2} \right)={{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}$.
We have ${{\cos }^{2}}\left( \dfrac{A-B}{2} \right)=\dfrac{1}{3}$.
Let us multiply both sides with ‘2’.
We have $2{{\cos }^{2}}\left( \dfrac{A-B}{2} \right)=2\times \dfrac{1}{3}$.
We have $2{{\cos }^{2}}\left( \dfrac{A-B}{2} \right)=\dfrac{2}{3}$.
Let us subtract both sides with ‘1’.
We have \[2{{\cos }^{2}}\left( \dfrac{A-B}{2} \right)-1=\dfrac{2}{3}-1\].
We have $2{{\cos }^{2}}\left( \dfrac{A-B}{2} \right)-1=\dfrac{-1}{3}$.
We know that $\cos \left( 2\theta \right)=2{{\cos }^{2}}\left( \theta \right)-1$.
We have $\cos \left( 2\times \dfrac{A-B}{2} \right)=\dfrac{-1}{3}$.
We have $\cos \left( A-B \right)=\dfrac{-1}{3}$.
$\therefore$ The value of $\cos \left( A-B \right)=\dfrac{-1}{3}$ ------(3).
Now we need to find the value of $\left| \cos A-\cos B \right|$ to check the remaining options.
We have $\left| \cos A-\cos B \right|$ and from trigonometric transformations, we have $\cos A-\cos B=-2.\sin \left( \dfrac{A+B}{2} \right).\sin \left( \dfrac{A-B}{2} \right)$.
We have $\left| \cos A-\cos B \right|=\left| -2.\sin \left( \dfrac{A+B}{2} \right).\sin \left( \dfrac{A-B}{2} \right) \right|$ -------(4).
Let us find the value of $\sin \left( \dfrac{A-B}{2} \right)$ first. From equation (2), we have got ${{\cos }^{2}}\left( \dfrac{A-B}{2} \right)=\dfrac{1}{3}$.
For any value of $\theta $, we have ${{\sin }^{2}}\left( \theta \right)=1-{{\cos }^{2}}\left( \theta \right)$.
So, we have ${{\sin }^{2}}\left( \dfrac{A-B}{2} \right)=1-{{\cos }^{2}}\left( \dfrac{A-B}{2} \right)$.
${{\sin }^{2}}\left( \dfrac{A-B}{2} \right)=1-\dfrac{1}{3}$.
${{\sin }^{2}}\left( \dfrac{A-B}{2} \right)=\dfrac{2}{3}$.
$\sin \left( \dfrac{A-B}{2} \right)=\sqrt{\dfrac{2}{3}}$.
We use $\sin \left( \dfrac{A-B}{2} \right)=\sqrt{\dfrac{2}{3}}$ and $A+B=\dfrac{\pi }{3}$ in equation (4).
$\left| \cos A-\cos B \right|=\left| -2.\sin \left( \dfrac{\dfrac{\pi }{3}}{2} \right).\sqrt{\dfrac{2}{3}} \right|$.
$\left| \cos A-\cos B \right|=\left| -2.\sin \left( \dfrac{\pi }{6} \right).\sqrt{\dfrac{2}{3}} \right|$.
We know that $\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}$.
$\left| \cos A-\cos B \right|=\left| -2.\left( \dfrac{1}{2} \right).\left( \sqrt{\dfrac{2}{3}} \right) \right|$.
$\left| \cos A-\cos B \right|=\left| -\sqrt{\dfrac{2}{3}} \right|$.
We know that $\left| x \right|>0$.
So, we get $\left| \cos A-\cos B \right|=\sqrt{\dfrac{2}{3}}$ -------(5).
From equations (3) and (5), we got $\cos \left( A-B \right)=\dfrac{-1}{3}$ and $\left| \cos A-\cos B \right|=\sqrt{\dfrac{2}{3}}$.
$\therefore$ The correct options are (b) and (c).
Note: Whenever we get problems that involve the sum of two trigonometric functions each of different angles, we start by using trigonometric transformation. We should use trigonometric identities and conversions of $\theta $ to $2\theta $, wherever required throughout the problem.
Complete step-by-step solution:
We have given that the values are $A+B=\dfrac{\pi }{3}$ and $\cos A+\cos B=1$. We need to check which are the true among the given options.
Let us keep $A+B=\dfrac{\pi }{3}$ as equation (1).
So, we have $\cos A+\cos B=1$.
From trigonometric transformations we know that $\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right).\cos \left( \dfrac{A-B}{2} \right)$.
We have $2\cos \left( \dfrac{A+B}{2} \right).\cos \left( \dfrac{A-B}{2} \right)=1$.
From the equation we have a value $A+B=\dfrac{\pi }{3}$, we substitute this in the above equation.
We have $2\cos \left( \dfrac{\dfrac{\pi }{3}}{2} \right).\cos \left( \dfrac{A-B}{2} \right)=1$.
We have $2\cos \left( \dfrac{\pi }{6} \right).\cos \left( \dfrac{A-B}{2} \right)=1$.
We know the value of $\cos \left( \dfrac{\pi }{6} \right)=\dfrac{\sqrt{3}}{2}$.
We have $2.\left( \dfrac{\sqrt{3}}{2} \right).\cos \left( \dfrac{A-B}{2} \right)=1$.
We have $\sqrt{3}.\cos \left( \dfrac{A-B}{2} \right)=1$.
We have $\cos \left( \dfrac{A-B}{2} \right)=\dfrac{1}{\sqrt{3}}$ -------(2).
Let us square equation (2) on both sides.
We have ${{\cos }^{2}}\left( \dfrac{A-B}{2} \right)={{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}$.
We have ${{\cos }^{2}}\left( \dfrac{A-B}{2} \right)=\dfrac{1}{3}$.
Let us multiply both sides with ‘2’.
We have $2{{\cos }^{2}}\left( \dfrac{A-B}{2} \right)=2\times \dfrac{1}{3}$.
We have $2{{\cos }^{2}}\left( \dfrac{A-B}{2} \right)=\dfrac{2}{3}$.
Let us subtract both sides with ‘1’.
We have \[2{{\cos }^{2}}\left( \dfrac{A-B}{2} \right)-1=\dfrac{2}{3}-1\].
We have $2{{\cos }^{2}}\left( \dfrac{A-B}{2} \right)-1=\dfrac{-1}{3}$.
We know that $\cos \left( 2\theta \right)=2{{\cos }^{2}}\left( \theta \right)-1$.
We have $\cos \left( 2\times \dfrac{A-B}{2} \right)=\dfrac{-1}{3}$.
We have $\cos \left( A-B \right)=\dfrac{-1}{3}$.
$\therefore$ The value of $\cos \left( A-B \right)=\dfrac{-1}{3}$ ------(3).
Now we need to find the value of $\left| \cos A-\cos B \right|$ to check the remaining options.
We have $\left| \cos A-\cos B \right|$ and from trigonometric transformations, we have $\cos A-\cos B=-2.\sin \left( \dfrac{A+B}{2} \right).\sin \left( \dfrac{A-B}{2} \right)$.
We have $\left| \cos A-\cos B \right|=\left| -2.\sin \left( \dfrac{A+B}{2} \right).\sin \left( \dfrac{A-B}{2} \right) \right|$ -------(4).
Let us find the value of $\sin \left( \dfrac{A-B}{2} \right)$ first. From equation (2), we have got ${{\cos }^{2}}\left( \dfrac{A-B}{2} \right)=\dfrac{1}{3}$.
For any value of $\theta $, we have ${{\sin }^{2}}\left( \theta \right)=1-{{\cos }^{2}}\left( \theta \right)$.
So, we have ${{\sin }^{2}}\left( \dfrac{A-B}{2} \right)=1-{{\cos }^{2}}\left( \dfrac{A-B}{2} \right)$.
${{\sin }^{2}}\left( \dfrac{A-B}{2} \right)=1-\dfrac{1}{3}$.
${{\sin }^{2}}\left( \dfrac{A-B}{2} \right)=\dfrac{2}{3}$.
$\sin \left( \dfrac{A-B}{2} \right)=\sqrt{\dfrac{2}{3}}$.
We use $\sin \left( \dfrac{A-B}{2} \right)=\sqrt{\dfrac{2}{3}}$ and $A+B=\dfrac{\pi }{3}$ in equation (4).
$\left| \cos A-\cos B \right|=\left| -2.\sin \left( \dfrac{\dfrac{\pi }{3}}{2} \right).\sqrt{\dfrac{2}{3}} \right|$.
$\left| \cos A-\cos B \right|=\left| -2.\sin \left( \dfrac{\pi }{6} \right).\sqrt{\dfrac{2}{3}} \right|$.
We know that $\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}$.
$\left| \cos A-\cos B \right|=\left| -2.\left( \dfrac{1}{2} \right).\left( \sqrt{\dfrac{2}{3}} \right) \right|$.
$\left| \cos A-\cos B \right|=\left| -\sqrt{\dfrac{2}{3}} \right|$.
We know that $\left| x \right|>0$.
So, we get $\left| \cos A-\cos B \right|=\sqrt{\dfrac{2}{3}}$ -------(5).
From equations (3) and (5), we got $\cos \left( A-B \right)=\dfrac{-1}{3}$ and $\left| \cos A-\cos B \right|=\sqrt{\dfrac{2}{3}}$.
$\therefore$ The correct options are (b) and (c).
Note: Whenever we get problems that involve the sum of two trigonometric functions each of different angles, we start by using trigonometric transformation. We should use trigonometric identities and conversions of $\theta $ to $2\theta $, wherever required throughout the problem.
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