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Question: How do you simplify \(\sin (ta{n^{ - 1}}x)\)?...

How do you simplify sin⁡(tan−1x)\sin (ta{n^{ - 1}}x)?

Explanation

Solution

Here basically we need to know that inverse of the trigonometric function can also be represented in the form by using the prefix arcarc Here we need to proceed by letting the value inside the bracket which is tan⁡−1x{\tan ^{ - 1}}x to be any variable say aa and then we will get tan⁡a=x\tan a = x and now we can easily find the value of sin⁡a\sin a which is required by using Pythagoras theorem.

Complete step by step solution:
Here we are given to simplify the term which is given as sin⁡(tan−1x)\sin (ta{n^{ - 1}}x)
So let us consider the term inside the bracket which is tan⁡−1x{\tan ^{ - 1}}x to be any variable say aa
So we get tan⁡−1x=a{\tan ^{ - 1}}x = a
So we will get tan⁡a=x\tan a = x
We can also write it as tan⁡a=x1\tan a = \dfrac{x}{1}
Now we need to find the value of sin⁡(tan−1x)\sin (ta{n^{ - 1}}x) which can be written as cos⁡a\cos a according to the variable which we have let tan⁡−1x=a{\tan ^{ - 1}}x = a
So let us consider the triangle ABCABC in which we can let the angle aa as ∠A\angle A and it is right angles at BB

Now we are given:
tan⁡a=x1\tan a = \dfrac{x}{1} −−−−(1) - - - - (1)
So we know that tan⁡a=perpendicularbase=BCAB\tan a = \dfrac{{{\text{perpendicular}}}}{{{\text{base}}}} = \dfrac{{BC}}{{AB}} −−−−−(2) - - - - - (2)
So by comparing the equation (1) and (2) we will get:
BC=x AB=1  BC = x \\\ AB = 1 \\\
Now we know that by Pythagoras theorem we can say that in the right angles triangle:
AC2=AB2+BC2A{C^2} = A{B^2} + B{C^2}
Now we can put in it:
BC=x AB=1  BC = x \\\ AB = 1 \\\
We will get:
AC2=AB2+BC2 AC2=12+x2 AC=1+x2  A{C^2} = A{B^2} + B{C^2} \\\ A{C^2} = {1^2} + {x^2} \\\ AC = \sqrt {1 + {x^2}} \\\
Now we know that in the right angles triangle:
sin⁡a=perpendicularhypotenuse=BCAC\sin a = \dfrac{{{\text{perpendicular}}}}{{{\text{hypotenuse}}}} = \dfrac{{BC}}{{AC}}
Now we can substitute the values of AB,ACAB,AC in the above equation of cos⁡a\cos a
sin⁡a=perpendicularhypotenuse=BCAC\sin a = \dfrac{{{\text{perpendicular}}}}{{{\text{hypotenuse}}}} = \dfrac{{BC}}{{AC}}
sin⁡a=x1+x2\sin a = \dfrac{x}{{\sqrt {1 + {x^2}} }}
Now we can substitute the value of aa and get:
sin⁡(tan−1x)\sin (ta{n^{ - 1}}x) =x1+x2 = \dfrac{x}{{\sqrt {1 + {x^2}} }}
So we can write sin⁡(tan⁡−1x)=sin⁡(arc⁡(tanx))\sin \left( {{{\tan }^{ - 1}}x} \right) = \sin (\operatorname{arc} (tanx)) =x1+x2 = \dfrac{x}{{\sqrt {1 + {x^2}} }}
Hence in this way by the use of Pythagoras theorem we can easily solve for such types of problems where we need to find the trigonometric function of the inverse function.

Note:
In such types of problems the student must keep in mind the basic trigonometric formula and the properties and also the use of Pythagoras theorem. We must know that cos⁡(sin⁡−1x)=sin⁡(cos⁡−1x)\cos \left( {{{\sin }^{ - 1}}x} \right) = \sin \left( {{{\cos }^{ - 1}}x} \right) because:
cos⁡(π2−cos⁡−1x)=sin⁡(cos⁡−1x)\cos \left( {\dfrac{\pi }{2} - {{\cos }^{ - 1}}x} \right) = \sin \left( {{{\cos }^{ - 1}}x} \right)