Question

Prove that \[{{\sin }^{-1}}\left( \dfrac{4}{5} \right)+{{\sin }^{-1}}\left( \dfrac{5}{13} \right)+{{\sin }^{-1}}\left( \dfrac{16}{65} \right)=\dfrac{\pi }{2}\].

Answer 1

Hint: We will begin with the left hand side of the given expression and then we will first apply the formula \[{{\sin }^{-1}}x+{{\sin }^{-1}}y={{\sin }^{-1}}(x\sqrt{1-{{y}^{2}}}+y\sqrt{1-{{x}^{2}}})\] on the first two terms. Also we will use the formula \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] to get sin in terms of cos.

Complete step-by-step answer:
Left hand side of the given expression is \[{{\sin }^{-1}}\left( \dfrac{4}{5} \right)+{{\sin }^{-1}}\left( \dfrac{5}{13} \right)+{{\sin }^{-1}}\left( \dfrac{16}{65} \right)........(1)\]
Now we know the formula that \[{{\sin }^{-1}}x+{{\sin }^{-1}}y={{\sin }^{-1}}(x\sqrt{1-{{y}^{2}}}+y\sqrt{1-{{x}^{2}}})\]. So applying this formula to the first two terms in equation (1) we get,
\[\Rightarrow {{\sin }^{-1}}\left( \dfrac{4}{5}\sqrt{1-{{\left( \dfrac{5}{13} \right)}^{2}}}+\dfrac{5}{13}\sqrt{1-{{\left( \dfrac{4}{5} \right)}^{2}}} \right)+{{\sin }^{-1}}\left( \dfrac{16}{65} \right)........(2)\]
Now squaring the terms in equation (2) we get,
\[\Rightarrow {{\sin }^{-1}}\left( \dfrac{4}{5}\sqrt{1-\dfrac{25}{169}}+\dfrac{5}{13}\sqrt{1-\dfrac{16}{25}} \right)+{{\sin }^{-1}}\left( \dfrac{16}{65} \right)........(3)\]
Now taking the LCM and simplifying the terms inside the root in equation (3) we get,
\[\begin{align}
  & \Rightarrow {{\sin }^{-1}}\left( \dfrac{4}{5}\sqrt{\dfrac{169-25}{169}}+\dfrac{5}{13}\sqrt{\dfrac{25-16}{25}} \right)+{{\sin }^{-1}}\left( \dfrac{16}{65} \right) \\
 & \Rightarrow {{\sin }^{-1}}\left( \dfrac{4}{5}\sqrt{\dfrac{144}{169}}+\dfrac{5}{13}\sqrt{\dfrac{9}{25}} \right)+{{\sin }^{-1}}\left( \dfrac{16}{65} \right)..........(4) \\
\end{align}\]
Now we can see that the numbers inside the root are perfect squares and hence we can simplify it in equation (4). So we get,
\[\Rightarrow {{\sin }^{-1}}\left( \dfrac{4}{5}\times \dfrac{12}{13}+\dfrac{5}{13}\times \dfrac{3}{5} \right)+{{\sin }^{-1}}\left( \dfrac{16}{65} \right)..........(5)\]
Now multiplying and adding the terms in equation (5) we get,
\[\Rightarrow {{\sin }^{-1}}\left( \dfrac{48}{65}+\dfrac{15}{65} \right)+{{\sin }^{-1}}\left( \dfrac{16}{65} \right)..........(6)\]
Now as the denominators are same inside the bracket in the first term of equation (6) we can directly add and hence we get,
\[\Rightarrow {{\sin }^{-1}}\left( \dfrac{63}{65} \right)+{{\sin }^{-1}}\left( \dfrac{16}{65} \right)..........(7)\]
Now let the first term of equation (7) be equal to \[\theta \] that is \[{{\sin }^{-1}}\left( \dfrac{63}{65} \right)=\theta ......(8)\].
Now rearranging sin in equation (8) we get,
\[\Rightarrow \sin \theta =\dfrac{63}{65}......(9)\]
Now squaring both sides of equation (9) we get,
\[\Rightarrow {{\sin }^{2}}\theta ={{\left( \dfrac{63}{65} \right)}^{2}}......(10)\]
Also we know that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]. So using this in equation (10) we get,
\[\Rightarrow 1-{{\cos }^{2}}\theta ={{\left( \dfrac{63}{65} \right)}^{2}}......(11)\]
Now solving for cos in equation (11) we get,
\[\begin{align}
  & \Rightarrow {{\cos }^{2}}\theta =1-{{\left( \dfrac{63}{65} \right)}^{2}} \\
 & \Rightarrow {{\cos }^{2}}\theta =\dfrac{{{65}^{2}}-{{63}^{2}}}{{{65}^{2}}}..........(12) \\
\end{align}\]
Now applying \[{{a}^{2}}-{{b}^{2}}=(a-b)(a+b)\] formula in equation (12) and solving we get,
\[\begin{align}
  & \Rightarrow {{\cos }^{2}}\theta =\dfrac{(65+63)(65-63)}{{{65}^{2}}} \\
 & \Rightarrow {{\cos }^{2}}\theta =\dfrac{128\times 2}{{{65}^{2}}} \\
 & \Rightarrow {{\cos }^{2}}\theta =\dfrac{256}{{{65}^{2}}} \\
 & \Rightarrow \cos \theta =\dfrac{16}{65}.........(13) \\
\end{align}\]
Now solving for \[\theta \] in equation (13) we get,
\[\Rightarrow \theta ={{\cos }^{-1}}\dfrac{16}{65}.........(14)\]
Now from equation (8) and equation (14) in equation (7) we get,
\[\Rightarrow {{\cos }^{-1}}\left( \dfrac{16}{65} \right)+{{\sin }^{-1}}\left( \dfrac{16}{65} \right).........(15)\]
And we know that \[{{\cos }^{-1}}A+{{\sin }^{-1}}A=\dfrac{\pi }{2}\] and hence applying this in equation (15) we get,
\[\Rightarrow {{\cos }^{-1}}\left( \dfrac{16}{65} \right)+{{\sin }^{-1}}\left( \dfrac{16}{65} \right)=\dfrac{\pi }{2}\]
Hence we have proved the left hand side of the given expression equal to the right hand side.

Note: Remembering the properties and formulas of inverse trigonometric equations is the key here. We in a hurry can make a mistake in applying the formula in equation (2) so we need to be very careful while doing this step. Also \[{{\cos }^{-1}}A+{{\sin }^{-1}}A=\dfrac{\pi }{2}\] is a very important property.