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**Hint:**For the problems of these kind where trigonometric and inverse trigonometry are involved we should use the trigonometry and inverse trigonometry properties like \[{{\sin }^{2}}t+{{\cos }^{2}}t=1\] and solve the questions. Here in the question we are asked to find the exact value of the \[\sin \left( {{\cos }^{-1}}\left( -\dfrac{1}{4} \right) \right)\]. So with the help of the quadrant system and above mentioned property we proceed in solving the question.

**Complete step-by-step solution:**

Here firstly we will imagine that inverse of cosine function as t and proceed with our further calculation as follows.

\[\Rightarrow {{\cos }^{-1}}\left( -\dfrac{1}{4} \right)=t\]

Here after that, we will transform the inverse function into a trigonometric function and the equation will be reduced as.

\[\Rightarrow \left( -\dfrac{1}{4} \right)=\cos t\]

Since the t is in quadrant 1 or 2 we will be knowing that the \[\sin t\] will be positive. The problem becomes as follows.

\[\Rightarrow \left( -\dfrac{1}{4} \right)=\cos t\] ,\[\sin t>0\]

After getting this value we use the basic trigonometric property that is \[{{\sin }^{2}}t+{{\cos }^{2}}t=1\] in the following step and after performing the calculations we will get the value of \[\sin t\] and hence we get the required answer. So the equation after performing the basic trigonometric property will be reduced as follows.

\[\Rightarrow {{\sin }^{2}}t+\dfrac{1}{16}=1\]

After sending the constants to the right-hand side of the equation the equation will be reduced as follows.

\[\Rightarrow {{\sin }^{2}}t=1-\dfrac{1}{16}\]

\[\Rightarrow {{\sin }^{2}}t=\dfrac{15}{16}\]

Here we will take the square root of the right-hand side so that we can obtain the required solution.

\[\Rightarrow \sin t=\dfrac{\sqrt{15}}{4}\]

**The above value is nothing but the required solution which is given below.**

\[\Rightarrow \sin \left( {{\cos }^{-1}}\left( -\dfrac{1}{4} \right) \right)=\dfrac{\sqrt{15}}{4}\]

\[\Rightarrow \sin \left( {{\cos }^{-1}}\left( -\dfrac{1}{4} \right) \right)=\dfrac{\sqrt{15}}{4}\]

**Note:**For solving these kinds of problems we must be having good knowledge in the basic trigonometry and its properties. In solving these types of questions we must be very careful in doing the calculations. We must also take care that this equation \[\Rightarrow {{\cos }^{-1}}\left( -\dfrac{1}{4} \right)=t\] can be transformed into \[\Rightarrow \left( -\dfrac{1}{4} \right)=\cos t\] if and only if this condition is satisfied \[\sin t> 0\] For solving these trigonometric questions we must its properties like \[{{\sin }^{2}}t+{{\cos }^{2}}t=1\] and also must be knowing the inverse trigonometric properties.

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