Solveeit Logo
Login

Question

Question: How do I evaluate \( \sin \left( {\arccos \left( {\dfrac{1}{3}} \right)} \right) \) ?...

How do I evaluate sin⁡(arccos⁡(13))\sin \left( {\arccos \left( {\dfrac{1}{3}} \right)} \right) ?

Explanation

Solution

Hint : In order to determine the simplification of the above question, let the arccos⁡(13)=θ\arccos \left( {\dfrac{1}{3}} \right) = \theta ,and transposing arccos⁡\arccos on the right-hand side. Now replacing the cos⁡θ\cos \theta with its value in the identity of trigonometry which state that sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 and after solving this for sin⁡θ\sin \theta ,put back the θ\theta assumed earlier to obtain your required result.
Formula :
sin⁡θ=tan⁡θsec⁡θ\sin \theta = \dfrac{{\tan \theta }}{{\sec \theta }}
sec⁡2θ=tan⁡2θ+1{\sec ^2}\theta = {\tan ^2}\theta + 1
sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1

Complete step-by-step answer :
We are given a trigonometric expression sin⁡(arccos⁡(13))\sin \left( {\arccos \left( {\dfrac{1}{3}} \right)} \right)
Let arccos⁡(13)=θ\arccos \left( {\dfrac{1}{3}} \right) = \theta ---------(1)
Transposing arccos⁡\arccos on the right hand side of the equation we get,
cos⁡θ=13\cos \theta = \dfrac{1}{3} ---------(2)
Since, we know the identity of trigonometry which states that the sum of square of sine and square of cosine is equal to one. i.e.
sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
Taking cos⁡2θ{\cos ^2}\theta on the right-hand side
sin⁡2θ=1−cos⁡2θ{\sin ^2}\theta = 1 - {\cos ^2}\theta
Now taking square root on both the sides, we get
sin⁡θ=±1−cos⁡2θ\sin \theta = \pm \sqrt {1 - {{\cos }^2}\theta }
Now putting the value of cos⁡θ\cos \theta from equation (2),
sin⁡θ=±1−(13)2 =±1−(19) =±1−19 =±9−19 =±89   \sin \theta = \pm \sqrt {1 - {{\left( {\dfrac{1}{3}} \right)}^2}} \\\ = \pm \sqrt {1 - \left( {\dfrac{1}{9}} \right)} \\\ = \pm \sqrt {1 - \dfrac{1}{9}} \\\ = \pm \sqrt {\dfrac{{9 - 1}}{9}} \\\ = \pm \sqrt {\dfrac{8}{9}} \;
Since 9=3\sqrt 9 = 3 and 8=4×2=22\sqrt 8 = \sqrt {4 \times 2} = 2\sqrt 2
sin⁡θ=±223\sin \theta = \pm \dfrac{{2\sqrt 2 }}{3}
Putting back the value of θ\theta ,which we have assumed in equation 1,
sin⁡(arccos⁡(13))=±223\sin \left( {\arccos \left( {\dfrac{1}{3}} \right)} \right) = \pm \dfrac{{2\sqrt 2 }}{3}
Therefore, the value of sin⁡(arccos⁡(13))\sin \left( {\arccos \left( {\dfrac{1}{3}} \right)} \right) is equal to ±223\pm \dfrac{{2\sqrt 2 }}{3} .
So, the correct answer is “ ±223\pm \dfrac{{2\sqrt 2 }}{3} ”.

Note : 1.One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.
2.Domain and range of inverse of cosine is [−1,1]\left[ { - 1,1} \right] and [0,π]\left[ {0,\pi } \right] respectively.
3. Domain and range of inverse of cosine is [−1,1]\left[ { - 1,1} \right] and [−π2,π2]\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] respectively.