If we have trigonometric ratios $\cos \left( \alpha +\beta \right)=\dfrac{4}{5},\sin \left( \alpha -\beta \right)=\dfrac{5}{13}$ and $\alpha ,\beta$ lie between 0 and $\dfrac{\pi }{4}$, find the value of $\tan 2\alpha$.
Answer
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Hint: Given that $\alpha ,\beta$ lie between 0 and $\dfrac{\pi }{4}$, then $\left( \alpha +\beta \right)$ lie between 0 and $\dfrac{\pi }{2}$. As $\sin \left( \alpha -\beta \right)=\dfrac{5}{13}$ and $\alpha ,\beta$ lie between 0 and $\dfrac{\pi }{4}$, $\alpha \gg \beta$ and $\left( \alpha -\beta \right)$ lie between 0 and $\dfrac{\pi }{4}$.
Using the relation ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we can find the values of $\cos \left( \alpha -\beta \right),\sin \left( \alpha +\beta \right)$. The required term is $\tan 2\alpha$. By writing $2\alpha$as$\left( \alpha +\beta \right)+\left( \alpha -\beta \right)$, we get $\tan 2\alpha$ as $\tan \left( \left( \alpha +\beta \right)+\left( \alpha -\beta \right) \right)$.
We know the formula of $\tan (A+B)=\dfrac{\tan A+\tan B}{1-\tan A\times \tan B}$. Using this formula we can get the answer.
Complete step-by-step solution:
In the question, it is given that $\cos \left( \alpha +\beta \right)=\dfrac{4}{5},\sin \left( \alpha -\beta \right)=\dfrac{5}{13}$.
We know the relation that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. Using this equation to get the values of $\cos \left( \alpha -\beta \right),\sin \left( \alpha +\beta \right)$.
\[\begin{align}
& {{\sin }^{2}}\left( \alpha +\beta \right)+{{\cos }^{2}}\left( \alpha +\beta \right)=1 \\
& {{\sin }^{2}}\left( \alpha +\beta \right)=1-{{\cos }^{2}}\left( \alpha +\beta \right) \\
& \sin \left( \alpha +\beta \right)=\sqrt{1-{{\cos }^{2}}\left( \alpha +\beta \right)} \\
\end{align}\]
Substituting the value of $\cos \left( \alpha +\beta \right)=\dfrac{4}{5}$ in the above equation we get,
\[\begin{align}
& \sin \left( \alpha +\beta \right)=\sqrt{1-{{\left( \dfrac{4}{5} \right)}^{2}}} \\
& \sin \left( \alpha +\beta \right)=\sqrt{1-\dfrac{16}{25}} \\
& \sin \left( \alpha +\beta \right)=\sqrt{\dfrac{25-16}{25}} \\
\end{align}\]
\[\sin \left( \alpha +\beta \right)=\sqrt{\dfrac{9}{25}}=\dfrac{3}{5}\]
We know that \[\tan \left( \alpha +\beta \right)=\dfrac{\sin \left( \alpha +\beta \right)}{\cos \left( \alpha +\beta \right)}\]
Substituting the values of \[\sin \left( \alpha +\beta \right)\] and \[\cos \left( \alpha +\beta \right)\] in the above equation, we get
\[\tan \left( \alpha +\beta \right)=\dfrac{\dfrac{3}{5}}{\dfrac{4}{5}}=\dfrac{3}{4}\to \left( 1 \right)\]
Similarly, for $\left( \alpha -\beta \right)$, we get
\[\begin{align}
& {{\sin }^{2}}\left( \alpha -\beta \right)+{{\cos }^{2}}\left( \alpha -\beta \right)=1 \\
& {{\cos }^{2}}\left( \alpha -\beta \right)=1-{{\sin }^{2}}\left( \alpha -\beta \right) \\
& {{\cos }^{2}}\left( \alpha -\beta \right)=\sqrt{1-{{\sin }^{2}}\left( \alpha -\beta \right)} \\
\end{align}\]
Substituting the value of $\sin \left( \alpha -\beta \right)=\dfrac{5}{13}$ in the above equation, we get
\[\begin{align}
& \cos \left( \alpha -\beta \right)=\sqrt{1-{{\left( \dfrac{5}{13} \right)}^{2}}} \\
& \cos \left( \alpha -\beta \right)=\sqrt{1-\dfrac{25}{169}} \\
& \cos \left( \alpha -\beta \right)=\sqrt{\dfrac{169-25}{169}} \\
\end{align}\]
\[\cos \left( \alpha -\beta \right)=\sqrt{\dfrac{144}{169}}=\dfrac{12}{13}\]
We know that \[\tan \left( \alpha -\beta \right)=\dfrac{\sin \left( \alpha -\beta \right)}{\cos \left( \alpha -\beta \right)}\]
Substituting the values of \[\sin \left( \alpha -\beta \right)\] and \[\cos \left( \alpha -\beta \right)\] in the above equation, we get
\[\tan \left( \alpha -\beta \right)=\dfrac{\dfrac{5}{13}}{\dfrac{12}{13}}=\dfrac{5}{12}\to \left( 2 \right)\]
The required term is $\tan 2\alpha$. To get the required term, we rewrite the expression as
$\tan 2\alpha =\tan \left( \left( \alpha +\beta \right)+\left( \alpha -\beta \right) \right)\to \left( 3 \right)$
We know the formula $\tan (A+B)=\dfrac{\tan A+\tan B}{1-\tan A\times \tan B}\to \left( 4 \right)$, using this formula in equation-3, we get
$\tan \left( \left( \alpha +\beta \right)+\left( \alpha -\beta \right) \right)=\dfrac{\tan \left( \alpha +\beta \right)+\tan \left( \alpha -\beta \right)}{1-\tan \left( \alpha +\beta \right)\tan \left( \alpha -\beta \right)}$.
Using the values of $\tan \left( \alpha +\beta \right)\text{ and }\tan \left( \alpha -\beta \right)$from the equations - 1 and 2, we get,
$\tan \left( \left( \alpha +\beta \right)+\left( \alpha -\beta \right) \right)=\dfrac{\dfrac{3}{4}+\dfrac{5}{12}}{1-\dfrac{3}{4}\times \dfrac{5}{12}}$
By simplifying, we get
$\tan \left( \left( \alpha +\beta \right)+\left( \alpha -\beta \right) \right)=\dfrac{\dfrac{3\times 3+5}{12}}{\dfrac{4\times 12-15}{4\times 12}}=\dfrac{\dfrac{14}{12}}{\dfrac{48-15}{48}}=\dfrac{14}{12}\times \dfrac{48}{33}$
$\tan \left( \left( \alpha +\beta \right)+\left( \alpha -\beta \right) \right)=\dfrac{56}{33}$
$\therefore \tan 2\alpha =\tan \left( \left( \alpha +\beta \right)+\left( \alpha -\beta \right) \right)=\dfrac{56}{33}$.
Note: An alternate way to do it is to convert both the given data into single trigonometric function and applying the inverse trigonometric concepts in it. That is
$\begin{align}
& \cos \left( \alpha +\beta \right)=\dfrac{4}{5}\Rightarrow \sin \left( \alpha +\beta \right)=\sqrt{1-{{\cos }^{2}}\left( \alpha +\beta \right)} \\
& \Rightarrow \sin \left( \alpha +\beta \right)=\sqrt{1-\dfrac{16}{25}}=\dfrac{3}{5} \\
& \sin \left( \alpha -\beta \right)=\dfrac{5}{13} \\
\end{align}$
We can write that
$\begin{align}
& \alpha +\beta ={{\sin }^{-1}}\dfrac{3}{5} \\
& \alpha -\beta ={{\sin }^{-1}}\dfrac{5}{13} \\
\end{align}$
We know that ${{\sin }^{-1}}\dfrac{a}{b}={{\tan }^{-1}}\dfrac{a}{\sqrt{{{b}^{2}}-{{a}^{2}}}}$
From this, we can write
$\begin{align}
& \alpha +\beta ={{\sin }^{-1}}\dfrac{3}{5}={{\tan }^{-1}}\dfrac{3}{\sqrt{{{5}^{2}}-{{3}^{2}}}}={{\tan }^{-1}}\dfrac{3}{4} \\
& \alpha -\beta ={{\sin }^{-1}}\dfrac{5}{13}={{\tan }^{-1}}\dfrac{5}{\sqrt{{{13}^{2}}-{{5}^{2}}}}={{\tan }^{-1}}\dfrac{5}{12} \\
\end{align}$
Adding the two equations, we get
$2\alpha ={{\tan }^{-1}}\dfrac{3}{4}+{{\tan }^{-1}}\dfrac{5}{12}$
Applying Tan on both sides and using equation - 4, we get
$\tan 2\alpha =\tan \left( {{\tan }^{-1}}\dfrac{3}{4}+{{\tan }^{-1}}\dfrac{5}{12} \right)=\dfrac{\tan \left( {{\tan }^{-1}}\dfrac{3}{4} \right)+\tan \left( {{\tan }^{-1}}\dfrac{5}{12} \right)}{1-\tan \left( {{\tan }^{-1}}\dfrac{3}{4} \right)\times \tan \left( {{\tan }^{-1}}\dfrac{5}{12} \right)}=\dfrac{\dfrac{3}{4}+\dfrac{5}{12}}{1-\dfrac{3}{4}\times \dfrac{5}{12}}$
$\tan 2\alpha =\dfrac{56}{33}$ same answer as the one we got with the other process.
Using the relation ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we can find the values of $\cos \left( \alpha -\beta \right),\sin \left( \alpha +\beta \right)$. The required term is $\tan 2\alpha$. By writing $2\alpha$as$\left( \alpha +\beta \right)+\left( \alpha -\beta \right)$, we get $\tan 2\alpha$ as $\tan \left( \left( \alpha +\beta \right)+\left( \alpha -\beta \right) \right)$.
We know the formula of $\tan (A+B)=\dfrac{\tan A+\tan B}{1-\tan A\times \tan B}$. Using this formula we can get the answer.
Complete step-by-step solution:
In the question, it is given that $\cos \left( \alpha +\beta \right)=\dfrac{4}{5},\sin \left( \alpha -\beta \right)=\dfrac{5}{13}$.
We know the relation that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. Using this equation to get the values of $\cos \left( \alpha -\beta \right),\sin \left( \alpha +\beta \right)$.
\[\begin{align}
& {{\sin }^{2}}\left( \alpha +\beta \right)+{{\cos }^{2}}\left( \alpha +\beta \right)=1 \\
& {{\sin }^{2}}\left( \alpha +\beta \right)=1-{{\cos }^{2}}\left( \alpha +\beta \right) \\
& \sin \left( \alpha +\beta \right)=\sqrt{1-{{\cos }^{2}}\left( \alpha +\beta \right)} \\
\end{align}\]
Substituting the value of $\cos \left( \alpha +\beta \right)=\dfrac{4}{5}$ in the above equation we get,
\[\begin{align}
& \sin \left( \alpha +\beta \right)=\sqrt{1-{{\left( \dfrac{4}{5} \right)}^{2}}} \\
& \sin \left( \alpha +\beta \right)=\sqrt{1-\dfrac{16}{25}} \\
& \sin \left( \alpha +\beta \right)=\sqrt{\dfrac{25-16}{25}} \\
\end{align}\]
\[\sin \left( \alpha +\beta \right)=\sqrt{\dfrac{9}{25}}=\dfrac{3}{5}\]
We know that \[\tan \left( \alpha +\beta \right)=\dfrac{\sin \left( \alpha +\beta \right)}{\cos \left( \alpha +\beta \right)}\]
Substituting the values of \[\sin \left( \alpha +\beta \right)\] and \[\cos \left( \alpha +\beta \right)\] in the above equation, we get
\[\tan \left( \alpha +\beta \right)=\dfrac{\dfrac{3}{5}}{\dfrac{4}{5}}=\dfrac{3}{4}\to \left( 1 \right)\]
Similarly, for $\left( \alpha -\beta \right)$, we get
\[\begin{align}
& {{\sin }^{2}}\left( \alpha -\beta \right)+{{\cos }^{2}}\left( \alpha -\beta \right)=1 \\
& {{\cos }^{2}}\left( \alpha -\beta \right)=1-{{\sin }^{2}}\left( \alpha -\beta \right) \\
& {{\cos }^{2}}\left( \alpha -\beta \right)=\sqrt{1-{{\sin }^{2}}\left( \alpha -\beta \right)} \\
\end{align}\]
Substituting the value of $\sin \left( \alpha -\beta \right)=\dfrac{5}{13}$ in the above equation, we get
\[\begin{align}
& \cos \left( \alpha -\beta \right)=\sqrt{1-{{\left( \dfrac{5}{13} \right)}^{2}}} \\
& \cos \left( \alpha -\beta \right)=\sqrt{1-\dfrac{25}{169}} \\
& \cos \left( \alpha -\beta \right)=\sqrt{\dfrac{169-25}{169}} \\
\end{align}\]
\[\cos \left( \alpha -\beta \right)=\sqrt{\dfrac{144}{169}}=\dfrac{12}{13}\]
We know that \[\tan \left( \alpha -\beta \right)=\dfrac{\sin \left( \alpha -\beta \right)}{\cos \left( \alpha -\beta \right)}\]
Substituting the values of \[\sin \left( \alpha -\beta \right)\] and \[\cos \left( \alpha -\beta \right)\] in the above equation, we get
\[\tan \left( \alpha -\beta \right)=\dfrac{\dfrac{5}{13}}{\dfrac{12}{13}}=\dfrac{5}{12}\to \left( 2 \right)\]
The required term is $\tan 2\alpha$. To get the required term, we rewrite the expression as
$\tan 2\alpha =\tan \left( \left( \alpha +\beta \right)+\left( \alpha -\beta \right) \right)\to \left( 3 \right)$
We know the formula $\tan (A+B)=\dfrac{\tan A+\tan B}{1-\tan A\times \tan B}\to \left( 4 \right)$, using this formula in equation-3, we get
$\tan \left( \left( \alpha +\beta \right)+\left( \alpha -\beta \right) \right)=\dfrac{\tan \left( \alpha +\beta \right)+\tan \left( \alpha -\beta \right)}{1-\tan \left( \alpha +\beta \right)\tan \left( \alpha -\beta \right)}$.
Using the values of $\tan \left( \alpha +\beta \right)\text{ and }\tan \left( \alpha -\beta \right)$from the equations - 1 and 2, we get,
$\tan \left( \left( \alpha +\beta \right)+\left( \alpha -\beta \right) \right)=\dfrac{\dfrac{3}{4}+\dfrac{5}{12}}{1-\dfrac{3}{4}\times \dfrac{5}{12}}$
By simplifying, we get
$\tan \left( \left( \alpha +\beta \right)+\left( \alpha -\beta \right) \right)=\dfrac{\dfrac{3\times 3+5}{12}}{\dfrac{4\times 12-15}{4\times 12}}=\dfrac{\dfrac{14}{12}}{\dfrac{48-15}{48}}=\dfrac{14}{12}\times \dfrac{48}{33}$
$\tan \left( \left( \alpha +\beta \right)+\left( \alpha -\beta \right) \right)=\dfrac{56}{33}$
$\therefore \tan 2\alpha =\tan \left( \left( \alpha +\beta \right)+\left( \alpha -\beta \right) \right)=\dfrac{56}{33}$.
Note: An alternate way to do it is to convert both the given data into single trigonometric function and applying the inverse trigonometric concepts in it. That is
$\begin{align}
& \cos \left( \alpha +\beta \right)=\dfrac{4}{5}\Rightarrow \sin \left( \alpha +\beta \right)=\sqrt{1-{{\cos }^{2}}\left( \alpha +\beta \right)} \\
& \Rightarrow \sin \left( \alpha +\beta \right)=\sqrt{1-\dfrac{16}{25}}=\dfrac{3}{5} \\
& \sin \left( \alpha -\beta \right)=\dfrac{5}{13} \\
\end{align}$
We can write that
$\begin{align}
& \alpha +\beta ={{\sin }^{-1}}\dfrac{3}{5} \\
& \alpha -\beta ={{\sin }^{-1}}\dfrac{5}{13} \\
\end{align}$
We know that ${{\sin }^{-1}}\dfrac{a}{b}={{\tan }^{-1}}\dfrac{a}{\sqrt{{{b}^{2}}-{{a}^{2}}}}$
From this, we can write
$\begin{align}
& \alpha +\beta ={{\sin }^{-1}}\dfrac{3}{5}={{\tan }^{-1}}\dfrac{3}{\sqrt{{{5}^{2}}-{{3}^{2}}}}={{\tan }^{-1}}\dfrac{3}{4} \\
& \alpha -\beta ={{\sin }^{-1}}\dfrac{5}{13}={{\tan }^{-1}}\dfrac{5}{\sqrt{{{13}^{2}}-{{5}^{2}}}}={{\tan }^{-1}}\dfrac{5}{12} \\
\end{align}$
Adding the two equations, we get
$2\alpha ={{\tan }^{-1}}\dfrac{3}{4}+{{\tan }^{-1}}\dfrac{5}{12}$
Applying Tan on both sides and using equation - 4, we get
$\tan 2\alpha =\tan \left( {{\tan }^{-1}}\dfrac{3}{4}+{{\tan }^{-1}}\dfrac{5}{12} \right)=\dfrac{\tan \left( {{\tan }^{-1}}\dfrac{3}{4} \right)+\tan \left( {{\tan }^{-1}}\dfrac{5}{12} \right)}{1-\tan \left( {{\tan }^{-1}}\dfrac{3}{4} \right)\times \tan \left( {{\tan }^{-1}}\dfrac{5}{12} \right)}=\dfrac{\dfrac{3}{4}+\dfrac{5}{12}}{1-\dfrac{3}{4}\times \dfrac{5}{12}}$
$\tan 2\alpha =\dfrac{56}{33}$ same answer as the one we got with the other process.
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