Question

Evaluate \[\sin \left[ {\dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{4}{5}} \right)} \right] = \]
A. \[\dfrac{1}{{\sqrt {10} }}\]
B. \[ - \dfrac{1}{{\sqrt {10} }}\]
C. \[\dfrac{1}{{10}}\]
D. \[ - \dfrac{1}{{10}}\]

Answer 1

Hint: In the given question, we need to find the value of \[\sin \left[ {\dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{4}{5}} \right)} \right]\]. For that, we know the formula \[{\cos ^{ - 1}}x = 2{\sin ^{ - 1}}\left( { \pm \sqrt {\dfrac{{1 - x}}{2}} } \right)\]. So, we apply this formula to our function and simplify it to get the desired result.

Formula used:
We have used the following formulas:
1. \[{\cos ^{ - 1}}x = 2{\sin ^{ - 1}}\left( { \pm \sqrt {\dfrac{{1 - x}}{2}} } \right)\]
2. \[\sin \left( {{{\sin }^{ - 1}}x} \right) = x\] as \[x \in \left[ { - 1,1} \right]\]

Complete step-by-step solution:
Given that \[\sin \left[ {\dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{4}{5}} \right)} \right]\]
We need to find the value of the given function.
Now we know that
\[{\cos ^{ - 1}}x = 2{\sin ^{ - 1}}\left( { \pm \sqrt {\dfrac{{1 - x}}{2}} } \right)\]
Now,by substituting this formula in our function, we get
\[
  \sin \left[ {\dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{4}{5}} \right)} \right] = \sin \left( {\dfrac{1}{2}2{{\sin }^{ - 1}}\left( { \pm \sqrt {\dfrac{{1 - \dfrac{4}{5}}}{2}} } \right)} \right) \\
   = \sin \left( {{{\sin }^{ - 1}}\left( { \pm \sqrt {\dfrac{{\dfrac{{5 - 4}}{5}}}{2}} } \right)} \right) \\
   = \sin \left( {{{\sin }^{ - 1}}\left( { \pm \sqrt {\dfrac{1}{{2 \times 5}}} } \right)} \right) \\
   = \sin \left( {{{\sin }^{ - 1}}\left( { \pm \sqrt {\dfrac{1}{{10}}} } \right)} \right)
 \]
On further simplification, we get
\[\sin \left[ {\dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{4}{5}} \right)} \right] = \sin \left( {{{\sin }^{ - 1}}\left( { \pm \dfrac{1}{{\sqrt {10} }}} \right)} \right)\]
Now we know that
\[\sin \left( {{{\sin }^{ - 1}}x} \right) = x\] as \[x \in \left[ { - 1,1} \right]\]
Therefore, \[\sin \left[ {\dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{4}{5}} \right)} \right] = \pm \dfrac{1}{{\sqrt {10} }}\]
Hence, option (A) and (B) is correct option.

Additional information: Inverse trigonometric functions perform the inverse operation of trigonometric functions like sine, cosine, tangent, cosecant, secant, and cotangent. We know that trig functions are particularly useful for right-angle triangles. The inverse trigonometric function is represented by the convention symbol, which includes arc-prefixes such as arc sin(x), arc cos(x), arc tan(x), arccsc(x), arcsec(x), and arc cot (x). The domain of the function is made up of all possible values of the independent variable where the function is defined, and the range is the set obtained by substituting all domain values into the function.

Note: Students should also use an alternate method to solve the above question, shown below:
 Given that \[\sin \left( {\dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{4}{5}} \right)} \right)\]
Now let us assume that \[{\cos ^{ - 1}}\left( {\dfrac{4}{5}} \right) = \theta \]
So, \[\cos \theta = \dfrac{4}{5}\]
Now we know that \[{\cos} \theta = 1 -2 {\sin ^2}\dfrac{\theta}{2} \]
\[
  1 - 2{\sin ^2}\dfrac{\theta }{2} = \dfrac{4}{5} \\
  2{\sin ^2}\dfrac{\theta }{2} = 1 - \dfrac{4}{5} \\
  2{\sin ^2}\dfrac{\theta }{2} = \dfrac{{5 - 4}}{5} \\
  2{\sin ^2}\dfrac{\theta }{2} = \dfrac{1}{5}
 \]
So,
\[
  1 - 2{\sin ^2}\dfrac{\theta }{2} = \dfrac{4}{5} \\
  2{\sin ^2}\dfrac{\theta }{2} = 1 - \dfrac{4}{5} \\
  2{\sin ^2}\dfrac{\theta }{2} = \dfrac{{5 - 4}}{5} \\
  2{\sin ^2}\dfrac{\theta }{2} = \dfrac{1}{5}
 \]
On further simplification, we get
\[
  {\sin ^2}\dfrac{\theta }{2} = \dfrac{1}{{5 \times 2}} \\
  {\sin ^2}\dfrac{\theta }{2} = \dfrac{1}{{10}} \\
  \sin \dfrac{\theta }{2} = \sqrt {\dfrac{1}{{10}}} \\
  \sin \dfrac{\theta }{2} = \pm \dfrac{1}{{\sqrt {10} }}
 \]
Therefore, \[\sin \left( {\dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{4}{5}} \right)} \right) = \pm \dfrac{1}{{\sqrt {10} }}\]
Hence, option (A) is correct option.