Solveeit Logo
Login

Question

Question: Evaluate \(\cos \left( {\arcsin \left( { - \dfrac{2}{3}} \right)} \right)\)....

Evaluate cos⁡(arcsin⁡(−23))\cos \left( {\arcsin \left( { - \dfrac{2}{3}} \right)} \right).

Explanation

Solution

We know that sin⁡−1x=arcsin⁡(x){\sin ^{ - 1}}x = \arcsin \left( x \right), such that here we have to evaluate in short cos⁡(sin⁡−1(−23))\cos \left( {{{\sin }^{ - 1}}\left( { - \dfrac{2}{3}} \right)} \right).
We also know the trigonometric identity sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 which can be used to evaluate the given question. So by using the above identities and information can evaluate the given question.

Complete step by step solution:
Given
cos⁡(arcsin⁡(−23))......................................(i)\cos \left( {\arcsin \left( { - \dfrac{2}{3}} \right)} \right)......................................\left( i \right)
Now we know to evaluate the value ofcos⁡(arcsin⁡(−23))\cos \left( {\arcsin \left( { - \dfrac{2}{3}} \right)} \right). For that purpose we can use various basic trigonometric identities such as sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos^2}\theta = 1.
Such that let:
p=arcsin⁡(−23) ⇒sin⁡p=(−23)........................(ii)  p = \arcsin \left( { - \dfrac{2}{3}} \right) \\\ \Rightarrow \sin p = \left( { - \dfrac{2}{3}} \right)........................\left( {ii} \right) \\\
Sincesin⁡−1x=arcsin⁡(x){\sin ^{ - 1}}x = \arcsin \left( x \right).
Now we have the identity sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1, from which we can find cos⁡p\cos p.
So on substituting the values in the identity sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 we can write:
cos⁡2p=1−sin⁡2p =1−(−23)2 =1−(49) =9−49 =59..................................(iii)  {\cos ^2}p = 1 - {\sin ^2}p \\\ = 1 - {\left( { - \dfrac{2}{3}} \right)^2} \\\ = 1 - \left( {\dfrac{4}{9}} \right) \\\ = \dfrac{{9 - 4}}{9} \\\ = \dfrac{5}{9}..................................\left( {iii} \right) \\\
So we get: cos⁡2p=59{\cos ^2}p = \dfrac{5}{9}
Now substituting back the value of ppwe can write:
cos⁡2(arcsin⁡(−23))=59 cos⁡(arcsin⁡(−23))=±53.........................(iv)  {\cos ^2}\left( {\arcsin \left( { - \dfrac{2}{3}} \right)} \right) = \dfrac{5}{9} \\\ \cos \left( {\arcsin \left( { - \dfrac{2}{3}} \right)} \right) = \pm \dfrac{{\sqrt 5 }}{3}.........................\left( {iv} \right) \\\
Such that there are two possibilities for the value of cos⁡(arcsin⁡(−23))\cos \left( {\arcsin \left( { - \dfrac{2}{3}} \right)} \right).
Now we also know that if:
arcsin⁡(x)=θ ⇒−π2⩽θ⩽π2.........................(v)  \arcsin \left( x \right) = \theta \\\ \Rightarrow - \dfrac{\pi }{2} \leqslant \theta \leqslant \dfrac{\pi }{2}.........................\left( v \right) \\\
So from the condition specified in (iv) we can take the positive value.
Therefore thee value ofcos⁡(arcsin⁡(−23))=53\cos \left( {\arcsin \left( { - \dfrac{2}{3}} \right)} \right) = \dfrac{{\sqrt 5 }}{3}.
Additional Information:
Some other properties useful for solving trigonometric questions:
Quadrant I: 0  −  π20\; - \;\dfrac{\pi }{2} All values are positive.
Quadrant II: π2  −  π\dfrac{\pi }{2}\; - \;\pi Only Sine and Cosec values are positive.
Quadrant III: π  −  3π2\pi \; - \;\dfrac{{3\pi }}{2} Only Tan and Cot values are positive.
Quadrant IV: 3π2  −  2π\dfrac{{3\pi }}{2}\; - \;2\pi Only Cos and Sec values are positive.

Note:
One of the basic property of arcsin⁡x\arcsin xis the range of the angle, which is given by:
arcsin⁡(x)=θ ⇒−π2⩽θ⩽π2  \arcsin \left( x \right) = \theta \\\ \Rightarrow - \dfrac{\pi }{2} \leqslant \theta \leqslant \dfrac{\pi }{2} \\\
So the values only in that range have to be taken under consideration.
Also while approaching a trigonometric problem one should keep in mind that one should work with one side at a time and manipulate it to the other side.