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Question: The value of \[\sin \left[ {2{{\cos }^{ - 1}}\dfrac{{\sqrt 5 }}{3}} \right]\]is A. \[\dfrac{{\sqrt...

The value of sin⁡[2cos⁡−153]\sin \left[ {2{{\cos }^{ - 1}}\dfrac{{\sqrt 5 }}{3}} \right]is
A. 53\dfrac{{\sqrt 5 }}{3}
B. 253\dfrac{{2\sqrt 5 }}{3}
C. 459\dfrac{{4\sqrt 5 }}{9}
D. 259\dfrac{{2\sqrt 5 }}{9}

Explanation

Solution

Use various trigonometric identities. Use Pythagoras Theorem (Pythagorean Theorem). Trigonometric and their inverse trigonometric parts get cancelled or compensated. Pythagoras theorem states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides“.

Complete step by step answer:
We need to find the value of sin⁡[2cos⁡−159]\sin \left[ {2{{\cos }^{ - 1}}\dfrac{{\sqrt 5 }}{9}} \right]
Using the property 2cos⁡−1x=cos⁡−1(2x2−1)2{\cos ^{ - 1}}x = {\cos ^{ - 1}}(2{x^2} - 1) we get
=sin⁡[cos⁡−1(2(53)2−1)]= \sin \left[ {{{\cos }^{ - 1}}\left( {2{{\left( {\dfrac{{\sqrt 5 }}{3}} \right)}^2} - 1} \right)} \right]
On simplifying the values in the angle part we get
=sin⁡[cos⁡−1(19)]= \sin \left[ {{{\cos }^{ - 1}}\left( {\dfrac{1}{9}} \right)} \right]
We can do this by using Pythagoras Theorem .
It is also sometimes called the Pythagorean Theorem. Pythagoras theorem states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides“. The sides of this triangle have been named as Perpendicular, Base and Hypotenuse. Here, the hypotenuse is the longest side, as it is opposite to the angle 90°. The sides of a right triangle (say a, b and c) which have positive integer values, when squared, are put into an equation, also called a Pythagorean triple.
Consider the triangle given :
Where “a” is the perpendicular,
“b” is the base,
“c” is the hypotenuse.
According to the definition, the Pythagoras Theorem formula is given as: (Hypotenuse)2=(Perpendicular)2+(Base)2{\left( {Hypotenuse} \right)^2} = {\left( {Perpendicular} \right)^2} + {\left( {Base} \right)^2}
That is (c)2=(a)2+(b)2{\left( c \right)^2} = {\left( a \right)^2} + {\left( b \right)^2}
Trigonometric functions in terms of sides of a right angled triangle:
sin⁡θ=ac\sin \theta = \dfrac{a}{c}
cos⁡θ=bc\cos \theta = \dfrac{b}{c}
tan⁡θ=ab\tan \theta = \dfrac{a}{b}
Changing cos⁡−1θ{\cos ^{ - 1}}\theta into sin⁡−1θ{\sin ^{ - 1}}\theta using the property that cos⁡−1θ=sin⁡−11−θ2{\cos ^{ - 1}}\theta = {\sin ^{ - 1}}\sqrt {1 - {\theta ^2}} we get
=sin⁡[sin⁡−11−(19)2]= \sin \left[ {{{\sin }^{ - 1}}\sqrt {1 - {{\left( {\dfrac{1}{9}} \right)}^2}} } \right]
This becomes
=sin⁡[sin⁡−18081]= \sin \left[ {{{\sin }^{ - 1}}\sqrt {\dfrac{{80}}{{81}}} } \right]
Now cancelling the trigonometric and inverse trigonometric part we get
=8081= \sqrt {\dfrac{{80}}{{81}}}
On simplifying the square root part we get
=809= \dfrac{{\sqrt {80} }}{9}
Which on further simplification becomes
=459= \dfrac{{4\sqrt 5 }}{9}

So, the correct answer is “Option C”.

Note: Use various trigonometric identities. Always try to make the trigonometric and its inverse part so as to ease the calculations. Trigonometric and their inverse trigonometric parts get cancelled or compensated.