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Question: Find the value of \[\sin \left[ \dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{4}{5} \right) \right]\]= ...

Find the value of sin⁡[12cos⁡−1(45)]\sin \left[ \dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{4}{5} \right) \right]=
1. 110\dfrac{1}{\sqrt{10}}
2. −110-\dfrac{1}{\sqrt{10}}
3. 110\dfrac{1}{10}
4. −110-\dfrac{1}{10}

Explanation

Solution

Hint : To solve this you must know the formula and properties of trigonometry. If you know them it is just a simple process of substituting the values in the formula giving the value needed. We also need to convert the given angle of cosine from its inverse form while making sure that it lies between the domain and range of these given expressions. After the simplification of the inverse function of cosine we just need to solve the value of sine function to get the answer we need for this question.

Complete step-by-step answer :
Consider the expression given in the problem it is
sin⁡[12cos⁡−1(45)]\sin \left[ \dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{4}{5} \right) \right]
Now to start simplifying this we need to know the value of the inverse of the cosine function here. We know that the principal value of inverse of cosine function as we know is:
cos⁡−1x ∈ (0,π){{\cos }^{-1}}x\text{ }\in \text{ }\left( 0,\pi \right) where the value of x ∈ [−1,1]x\text{ }\in \text{ }\left[ -1,1 \right]
Now as we know that 45\dfrac{4}{5}belongs in the range from −1,1-1,1 there the inverse of cosine function will exit which means that we can further solve this equation to get the answer we need here
Now to find the value of cosine we let
θ=cos⁡−1(45)\theta ={{\cos }^{-1}}\left( \dfrac{4}{5} \right)
Which means
cos⁡θ=45\cos \theta =\dfrac{4}{5}
Using this we can find the sine which will be
sin⁡θ=1−(45)2\sin \theta =\sqrt{1-{{\left( \dfrac{4}{5} \right)}^{2}}}
sin⁡θ=35\sin \theta =\dfrac{3}{5}
But since in the question when we apply the θ\theta we get that we need sin⁡[θ2]\sin \left[ \dfrac{\theta }{2} \right]we need to convert this in the form of half angle
cos⁡θ=1−2sin⁡2θ2\cos \theta =1-2{{\sin }^{2}}\dfrac{\theta }{2}
Hence substituting the values we get
sin⁡2θ2=1−cos⁡θ2{{\sin }^{2}}\dfrac{\theta }{2}=\dfrac{1-\cos \theta }{2}
sin⁡2θ2=1−452{{\sin }^{2}}\dfrac{\theta }{2}=\dfrac{1-\dfrac{4}{5}}{2}
sin⁡2θ2=110{{\sin }^{2}}\dfrac{\theta }{2}=\dfrac{1}{10}
sin⁡θ2=110\sin \dfrac{\theta }{2}=\dfrac{1}{\sqrt{10}}
θ2=sin⁡−1(110)\dfrac{\theta }{2}={{\sin }^{-1}}\left( \dfrac{1}{\sqrt{10}} \right)
Putting this value in our expression we get
sin⁡(sin⁡−1(110))\sin \left( {{\sin }^{-1}}\left( \dfrac{1}{\sqrt{10}} \right) \right)
This gives us the answer to the question which is 110\dfrac{1}{\sqrt{10}}. Hence the answer is option 1
So, the correct answer is “Option 1”.

Note : We should keep in mind while solving this question that the value inside the inverse of the trigonometric question must be within the range because if it isn’t it can change the answer accordingly. A common mistake made in questions like this could be that the student does not realize that the function is out of domain. Another mistake made is people try finding the angle of inverse function. In cases where the angle isn’t one of the known angle values then try simplifying it using different methods. Just like we used the supplementary angle logic here.