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Question: How do you evaluate \(\tan \left( \arcsin \left( \dfrac{1}{3} \right) \right)\)?...

How do you evaluate tan⁡(arcsin⁡(13))\tan \left( \arcsin \left( \dfrac{1}{3} \right) \right)?

Explanation

Solution

First express the expression in terms of sine and cosine form using the formula tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x}. Then simplify the numerator and the denominator by taking them separately. For numerator use the formula sin⁡(arcsin⁡(x))=x\sin \left( \arcsin \left( x \right) \right)=xand for the denominator use cos⁡x=1−sin⁡2x\cos x=\sqrt{1-{{\sin }^{2}}x} and sin⁡2(arcsin⁡(x))=x2{{\sin }^{2}}\left( \arcsin \left( x \right) \right)={{x}^{2}}. After getting simplified values, bring the numerator and the denominator together and do the necessary calculation to obtain the required solution.

Complete step by step solution:
As we know, tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x}
Now, considering our question tan⁡(arcsin⁡(13))\tan \left( \arcsin \left( \dfrac{1}{3} \right) \right)
It can be written as tan⁡(arcsin⁡(13))=sin⁡(arcsin⁡(13))cos⁡(arcsin⁡(13))\tan \left( \arcsin \left( \dfrac{1}{3} \right) \right)=\dfrac{\sin \left( \arcsin \left( \dfrac{1}{3} \right) \right)}{\cos \left( \arcsin \left( \dfrac{1}{3} \right) \right)}
From here, we have to simplify the numerator part and the denominator part separately.
For the numerator part:
As we know, sin⁡(arcsin⁡(x))=x\sin \left( \arcsin \left( x \right) \right)=x
So, sin⁡(arcsin⁡(13))\sin \left( \arcsin \left( \dfrac{1}{3} \right) \right) can be written as sin⁡(arcsin⁡(13))=13\sin \left( \arcsin \left( \dfrac{1}{3} \right) \right)=\dfrac{1}{3}
For the denominator part:
As we know
sin⁡2x+cos⁡2x=1 ⇒cos⁡2x=1−sin⁡2x ⇒cos⁡x=1−sin⁡2x \begin{aligned} & {{\sin }^{2}}x+{{\cos }^{2}}x=1 \\\ & \Rightarrow {{\cos }^{2}}x=1-{{\sin }^{2}}x \\\ & \Rightarrow \cos x=\sqrt{1-{{\sin }^{2}}x} \\\ \end{aligned}
So, cos⁡(arcsin⁡(13))\cos \left( \arcsin \left( \dfrac{1}{3} \right) \right) can be written as cos⁡(arcsin⁡(13))=1−sin⁡2(arcsin⁡(13))\cos \left( \arcsin \left( \dfrac{1}{3} \right) \right)=\sqrt{1-{{\sin }^{2}}\left( \arcsin \left( \dfrac{1}{3} \right) \right)}
Again, we know sin⁡2(arcsin⁡(x))=(x)2=x2{{\sin }^{2}}\left( \arcsin \left( x \right) \right)={{\left( x \right)}^{2}}={{x}^{2}}
So, sin⁡2(arcsin⁡(13))=(13)2=19{{\sin }^{2}}\left( \arcsin \left( \dfrac{1}{3} \right) \right)={{\left( \dfrac{1}{3} \right)}^{2}}=\dfrac{1}{9}
Putting the value of sin⁡2(arcsin⁡(13)){{\sin }^{2}}\left( \arcsin \left( \dfrac{1}{3} \right) \right)in cos⁡(arcsin⁡(13))=1−sin⁡2(arcsin⁡(13))\cos \left( \arcsin \left( \dfrac{1}{3} \right) \right)=\sqrt{1-{{\sin }^{2}}\left( \arcsin \left( \dfrac{1}{3} \right) \right)}, we get
cos⁡(arcsin⁡(13))=1−19=9−19=89=223\cos \left( \arcsin \left( \dfrac{1}{3} \right) \right)=\sqrt{1-\dfrac{1}{9}}=\sqrt{\dfrac{9-1}{9}}=\sqrt{\dfrac{8}{9}}=\dfrac{2\sqrt{2}}{3}
Bringing the numerator and the denominator together, now our expression becomes
tan⁡(arcsin⁡(13))=sin⁡(arcsin⁡(13))cos⁡(arcsin⁡(13))=13223=13×322=122\tan \left( \arcsin \left( \dfrac{1}{3} \right) \right)=\dfrac{\sin \left( \arcsin \left( \dfrac{1}{3} \right) \right)}{\cos \left( \arcsin \left( \dfrac{1}{3} \right) \right)}=\dfrac{\dfrac{1}{3}}{\dfrac{2\sqrt{2}}{3}}=\dfrac{1}{3}\times \dfrac{3}{2\sqrt{2}}=\dfrac{1}{2\sqrt{2}}
This is the required solution of the given question.

Note: We know, sin⁡2x{{\sin }^{2}}x can be written as (sin⁡x)2{{\left( \sin x \right)}^{2}}. Similarly we can write sin⁡2(arcsin⁡(x)){{\sin }^{2}}\left( \arcsin \left( x \right) \right) as [sin⁡(arcsin⁡(x))]2{{\left[ \sin \left( \arcsin \left( x \right) \right) \right]}^{2}}. Again as we know, the value of sin⁡(arcsin⁡(x))=x\sin \left( \arcsin \left( x \right) \right)=x so, the value of [sin⁡(arcsin⁡(x))]2=(x)2=x2{{\left[ \sin \left( \arcsin \left( x \right) \right) \right]}^{2}}={{\left( x \right)}^{2}}={{x}^{2}}, which is used in this question. The above question should be solved by taking the numerator and the denominator separately to avoid errors and complex calculations.