Question

How do you find the value of \[\sin (\arccos (x))\] ?

Answer 1

Hint: Here arccosine is used to denote the inverse of cosine which is represented as \[{{\cos }^{-1}}\] . The sine of any angle is the ratio of length of opposite side to the length of hypotenuse in a right-angle triangle. For example, graphically \[\sin \left( \dfrac{\pi }{2} \right)=1\] . Since here we have to find arccosine for a general value. Therefore, we will apply some basic trigonometric identities to find its value.

Complete step by step answer:
This question belongs to the concept of inverse trigonometric functions. In the question we have to find \[\sin (\arccos (x))\] . Arccosine or cosine inverse is inverse of a cosine expression which gives the value of angle. It is equal to the ratio of length of hypotenuse to the length of the perpendicular or base side in a right-angle triangle. For example, \[{{\cos }^{-1}}(1)=0\] .
Here we have \[\sin (\arccos (x))\] . In trigonometry we have one identity, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]
Also, if the range of x is \[[-1,1]\] and \[\theta =\arccos (x)\] then we can say that the range of \[\theta \] is \[[0,\pi ]\] .
From this we can conclude that \[\sin (\theta )\ge 0\] .
Hence,
\[\sin (\arccos (x))=\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }=\sqrt{1-{{x}^{2}}}\]
Therefore, the solution is \[\sqrt{1-{{x}^{2}}}\] .

Note: While solving the question of this type for non-transcendental values keep in mind the range and domain of the trigonometric functions. Keep in mind the identity used in the above question for future use. In the above question we Also, when we are finding values of trigonometric functions, it is better to use a graph of that function.