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Question: How do you find the value of \(\sin \left( {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) + {{\tan }...

How do you find the value of sin⁡(tan⁡−1(12)+tan⁡−1(13))\sin \left( {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) + {{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right)?

Explanation

Solution

To find the value of given trigonometric function we will use trigonometric identity tan⁡−1(x)+tan⁡−1(y)=tan⁡−1(x+y1−xy){\tan ^{ - 1}}(x) + {\tan ^{ - 1}}(y) = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right). After implementing this identity we will find the value of the sin⁡\sin function.

Complete step by step answer:
Here we are given to simplify the term sin⁡(tan⁡−1(12)+tan⁡−1(13))\sin \left( {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) + {{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right)
First, we solve (tan⁡−1(12)+tan⁡−1(13))\left( {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) + {{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right)
We know that
tan⁡−1(x)+tan⁡−1(y)=tan⁡−1(x+y1−xy){\tan ^{ - 1}}(x) + {\tan ^{ - 1}}(y) = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)
Putting x=12x = \dfrac{1}{2} and y=13y = \dfrac{1}{3} in the above formula. We get,
⇒\Rightarrow (tan⁡−1(12)+tan⁡−1(13))=tan⁡−1(12+131−12×13)\left( {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) + {{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right) = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{1}{2} + \dfrac{1}{3}}}{{1 - \dfrac{1}{2} \times \dfrac{1}{3}}}} \right)
Simplifying the above equation. We get,
⇒tan⁡−1(3+261−16)=tan⁡−1(566−16)\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{3 + 2}}{6}}}{{1 - \dfrac{1}{6}}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{5}{6}}}{{\dfrac{{6 - 1}}{6}}}} \right)
Further solving the above equation. We get,
⇒\Rightarrow tan⁡−1(5656){\tan ^{ - 1}}\left( {\dfrac{{\dfrac{5}{6}}}{{\dfrac{5}{6}}}} \right) =tan⁡−1(56×65) = {\tan ^{ - 1}}\left( {\dfrac{5}{6} \times \dfrac{6}{5}} \right)
⇒tan⁡−1(1)\Rightarrow {\tan ^{ - 1}}(1)
Let tan⁡−1(1)=A{\tan ^{ - 1}}(1) = A
So, tan⁡(A)=1\tan (A) = 1
We know that tan⁡(45∘)=1\tan (45^\circ ) = 1

Here, we get A=45∘A = 45^\circ
Therefore, tan⁡−1(1)=45∘{\tan ^{ - 1}}(1) = 45^\circ
So, (tan⁡−1(12)+tan⁡−1(13))\left( {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) + {{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right) =45∘= 45^\circ
Therefore,
⇒\Rightarrow sin⁡(tan⁡−1(12)+tan⁡−1(13))\sin \left( {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) + {{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right) =sin⁡(45∘) = \sin (45^\circ )
We know that sin⁡(45∘)=12\sin (45^\circ ) = \dfrac{1}{{\sqrt 2 }}
Therefore, we get
sin⁡(tan⁡−1(12)+tan⁡−1(13))=12\sin \left( {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) + {{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right) = \dfrac{1}{{\sqrt 2 }}
Hence, the value of sin⁡(tan⁡−1(12)+tan⁡−1(13))\sin \left( {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) + {{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right) is equal to 12\dfrac{1}{{\sqrt 2 }}.

Note: In this types of problems the student must remember the basic trigonometric identities and the value of angles of trigonometric function such as tan⁡(45∘)=1\tan (45^\circ ) = 1 and sin⁡(45∘)=12\sin (45^\circ ) = \dfrac{1}{{\sqrt 2 }}.
In these types of problems first reduce the function by using different trigonometric identities and then solve it. After reducing it will be easier to solve the problem.