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Question: Prove the following \({{\cos }^{-1}}\left( \dfrac{4}{5} \right)+{{\cos }^{-1}}\left( \dfrac{12}{13} ...

Prove the following cos⁡−1(45)+cos⁡−1(1213)=cos⁡−1(3365){{\cos }^{-1}}\left( \dfrac{4}{5} \right)+{{\cos }^{-1}}\left( \dfrac{12}{13} \right)={{\cos }^{-1}}\left( \dfrac{33}{65} \right) .

Explanation

Solution

Hint: In the given question we are asked to prove L.H.S = R.H.S. To prove that first we need to find the sum of cos⁡−1(45){{\cos }^{-1}}\left( \dfrac{4}{5} \right) and cos⁡−1(1213){{\cos }^{-1}}\left( \dfrac{12}{13} \right) with the help of formula i.e. cos⁡−1x+cos⁡−1y=cos⁡−1(xy−1−x21−y2){{\cos }^{-1}}x+{{\cos }^{-1}}y={{\cos }^{-1}}\left( xy-\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}} \right) to get the answer.

Complete step-by-step answer:
To prove that cos⁡−1(45)+cos⁡−1(1213)=cos⁡−1(3365){{\cos }^{-1}}\left( \dfrac{4}{5} \right)+{{\cos }^{-1}}\left( \dfrac{12}{13} \right)={{\cos }^{-1}}\left( \dfrac{33}{65} \right) let us take x=45x=\dfrac{4}{5} and y=1213y=\dfrac{12}{13} .
We know that cos⁡−1x+cos⁡−1y=cos⁡−1(xy−1−x21−y2){{\cos }^{-1}}x+{{\cos }^{-1}}y={{\cos }^{-1}}\left( xy-\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}} \right).
We have,
cos⁡−1(45)+cos⁡−1(1213){{\cos }^{-1}}\left( \dfrac{4}{5} \right)+{{\cos }^{-1}}\left( \dfrac{12}{13} \right).
=cos⁡−1(45×1213−1−(45)21−(1213)2)={{\cos }^{-1}}\left( \dfrac{4}{5}\times \dfrac{12}{13}-\sqrt{1-{{\left( \dfrac{4}{5} \right)}^{2}}}\sqrt{1-{{\left( \dfrac{12}{13} \right)}^{2}}} \right).
=cos⁡−1(4865−1−16251−144169)={{\cos }^{-1}}\left( \dfrac{48}{65}-\sqrt{1-\dfrac{16}{25}}\sqrt{1-\dfrac{144}{169}} \right) .
=cos⁡−1(4865−25−1625169−144169)={{\cos }^{-1}}\left( \dfrac{48}{65}-\sqrt{\dfrac{25-16}{25}}\sqrt{\dfrac{169-144}{169}} \right).
=cos⁡−1(4865−(35×513))={{\cos }^{-1}}\left( \dfrac{48}{65}-\left( \dfrac{3}{5}\times \dfrac{5}{13} \right) \right) .
=cos⁡−1(4865−1565)={{\cos }^{-1}}\left( \dfrac{48}{65}-\dfrac{15}{65} \right) .
=cos⁡−1(48−1565)={{\cos }^{-1}}\left( \dfrac{48-15}{65} \right) .
=cos⁡−1(3365)={{\cos }^{-1}}\left( \dfrac{33}{65} \right) .
Hence proved,
cos⁡−1(45)+cos⁡−1(1213)=cos⁡−1(3365){{\cos }^{-1}}\left( \dfrac{4}{5} \right)+{{\cos }^{-1}}\left( \dfrac{12}{13} \right)={{\cos }^{-1}}\left( \dfrac{33}{65} \right) .

Note: There’s an alternative way to solve this question,
Let a=cos⁡−145a={{\cos }^{-1}}\dfrac{4}{5} and b=cos⁡−11213b={{\cos }^{-1}}\dfrac{12}{13}
Let a=cos⁡−145a={{\cos }^{-1}}\dfrac{4}{5}
cos⁡a=45\cos a=\dfrac{4}{5}
We know that ,
sin⁡2a=1−cos⁡2a sin⁡a=1−cos⁡2a =1−(45)2=35 \begin{aligned} & {{\sin }^{2}}a=1-{{\cos }^{2}}a \\\ & \sin a=\sqrt{1-{{\cos }^{2}}a} \\\ & =\sqrt{1-{{\left( \dfrac{4}{5} \right)}^{2}}}=\dfrac{3}{5} \\\ \end{aligned}
Let b=cos⁡−11213b={{\cos }^{-1}}\dfrac{12}{13}
cos⁡b=1213\cos b=\dfrac{12}{13}
We know that ,
sin⁡2b=1−cos⁡2b sin⁡b=1−cos⁡2b =1−(1213)2=513 \begin{aligned} & {{\sin }^{2}}b=1-{{\cos }^{2}}b \\\ & \sin b=\sqrt{1-{{\cos }^{2}}b} \\\ & =\sqrt{1-{{\left( \dfrac{12}{13} \right)}^{2}}}=\dfrac{5}{13} \\\ \end{aligned}
We know that cos⁡(a+b)=cos⁡acos⁡−sin⁡asin⁡b\cos \left( a+b \right)=\cos a\cos -\sin a\sin b
Put the values cos⁡a=45,cos⁡b=1213,sin⁡a=35and sin⁡b=513\cos a=\dfrac{4}{5},\cos b=\dfrac{12}{13},\sin a=\dfrac{3}{5}\text{and }\sin b=\dfrac{5}{13} .
cos⁡(a+b)=45×1213−35×513 =48−1565=3365 ∴cos⁡(a+b)=3365 \begin{aligned} & \cos \left( a+b \right)=\dfrac{4}{5}\times \dfrac{12}{13}-\dfrac{3}{5}\times \dfrac{5}{13} \\\ & =\dfrac{48-15}{65}=\dfrac{33}{65} \\\ & \therefore \cos \left( a+b \right)=\dfrac{33}{65} \\\ \end{aligned}
a+b=cos⁡−1(3365)a+b={{\cos }^{-1}}\left( \dfrac{33}{65} \right)
cos⁡−1(45)+cos⁡−1(1213)=cos⁡−1(3365){{\cos }^{-1}}\left( \dfrac{4}{5} \right)+{{\cos }^{-1}}\left( \dfrac{12}{13} \right)={{\cos }^{-1}}\left( \dfrac{33}{65} \right)
Hence proved.