Question

Given $\csc x=8$; how do you find $\sin \dfrac{x}{2}$ , $\cos \dfrac{x}{2}$,$\tan \dfrac{x}{2}$?
(a) Using trigonometric identities
(b) Using triangles
(c) Using linear formulas
(d) All of them

Answer 1

Hint: We are to find the value of $\sin \dfrac{x}{2}$ , $\cos \dfrac{x}{2}$,$\tan \dfrac{x}{2}$where we were given $\csc x=8$. Now, if we also know $\cos x=\sqrt{1-{{\sin }^{2}}x}$, we will initially find cos x. Then we will use the formula of $\cos x=2{{\cos }^{2}}\dfrac{x}{2}-1$and ${{\sin }^{2}}\dfrac{x}{2}=\dfrac{1}{2}\left( 1-\cos x \right)$, to get the value of $\sin \dfrac{x}{2}$ , $\cos \dfrac{x}{2}$. Then using the value of $\tan x=\dfrac{\sin x}{\cos x}$ , we also get the value of $\tan \dfrac{x}{2}$.

Complete step by step solution:
We will start with,
$\csc x=8$;
We also know,$\sin x=\dfrac{1}{\csc x}=\dfrac{1}{8}$ ,
So, $\sin x=\dfrac{1}{8}$,
And as, ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ , we have, ${{\cos }^{2}}x=1-{{\sin }^{2}}x$
So, $\cos x=\sqrt{1-{{\sin }^{2}}x}$
Putting the value of sin x we get,
$\cos x=\sqrt{1-{{\left( \dfrac{1}{8} \right)}^{2}}}$
Simplifying,
$\cos x=\sqrt{1-\dfrac{1}{64}}=\sqrt{\dfrac{63}{64}}$
So, $\cos x=\pm \dfrac{\sqrt{63}}{8}$
On the other hand, we also know, $\sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$ and $\cos x=2{{\cos }^{2}}\dfrac{x}{2}-1$ , using the trigonometric identities.
We have, $\cos x=2{{\cos }^{2}}\dfrac{x}{2}-1$,
$\Rightarrow {{\cos }^{2}}\dfrac{x}{2}=\dfrac{1}{2}\left( 1+\cos x \right)$
Now, putting the value, $\cos x=\pm \dfrac{\sqrt{63}}{8}$,
$\Rightarrow {{\cos }^{2}}\dfrac{x}{2}=\dfrac{1}{2}\left( 1\pm \dfrac{\sqrt{63}}{8} \right)$
So, we are getting two values, ${{\cos }^{2}}\dfrac{x}{2}=\left( \dfrac{8\pm \sqrt{63}}{16} \right)$.
Hence,
$\cos \dfrac{x}{2}=\sqrt{\dfrac{8\pm \sqrt{63}}{16}}$
More simplifying,
$\cos \dfrac{x}{2}=\dfrac{\sqrt{8\pm 3\sqrt{7}}}{4}=\dfrac{1}{2}\left( 3\sqrt{2}\pm \sqrt{14} \right)$
Again, we also know, $\cos x=1-2{{\sin }^{2}}\dfrac{x}{2}$,
Then, ${{\sin }^{2}}\dfrac{x}{2}=\dfrac{1}{2}\left( 1-\cos x \right)$,
Now, putting the value, $\cos x=\pm \dfrac{\sqrt{63}}{8}$,
$\Rightarrow {{\sin }^{2}}\dfrac{x}{2}=\dfrac{1}{2}\left( 1\mp \dfrac{\sqrt{63}}{8} \right)$
So, we are getting two values, ${{\sin }^{2}}\dfrac{x}{2}=\left( \dfrac{8\mp \sqrt{63}}{16} \right)$.
Hence,
$\sin \dfrac{x}{2}=\sqrt{\dfrac{8\mp \sqrt{63}}{16}}$
More simplifying,
$\sin \dfrac{x}{2}=\dfrac{\sqrt{8\mp 3\sqrt{7}}}{4}=\dfrac{1}{2}\left( 3\sqrt{2}\mp \sqrt{14} \right)$
So, if $\cos \dfrac{x}{2}=\dfrac{1}{2}\left( 3\sqrt{2}+\sqrt{14} \right)$ and $\sin \dfrac{x}{2}=\dfrac{1}{2}\left( 3\sqrt{2}-\sqrt{14} \right)$and $\tan \dfrac{x}{2}=\dfrac{\dfrac{1}{2}\left( 3\sqrt{2}-\sqrt{14} \right)}{\dfrac{1}{2}\left( 3\sqrt{2}+\sqrt{14} \right)}$
Thus, $\tan \dfrac{x}{2}=\dfrac{3\sqrt{2}-\sqrt{14}}{3\sqrt{2}+\sqrt{14}}=\dfrac{{{\left( 3\sqrt{2}-\sqrt{14} \right)}^{2}}}{{{\left( 3\sqrt{2} \right)}^{2}}-{{\left( \sqrt{14} \right)}^{2}}}$
More simplifying, \[\tan \dfrac{x}{2}=\dfrac{{{\left( 3\sqrt{2} \right)}^{2}}+{{\left( \sqrt{14} \right)}^{2}}-2.3\sqrt{2}.\sqrt{14}}{18-14}=\dfrac{18+14-12\sqrt{7}}{4}=\dfrac{32-12\sqrt{7}}{4}=8-3\sqrt{7}\]
And, on the other hand,
If $\cos \dfrac{x}{2}=\dfrac{1}{2}\left( 3\sqrt{2}-\sqrt{14} \right)$ and $\sin \dfrac{x}{2}=\dfrac{1}{2}\left( 3\sqrt{2}+\sqrt{14} \right)$and $\tan \dfrac{x}{2}=\dfrac{\dfrac{1}{2}\left( 3\sqrt{2}+\sqrt{14} \right)}{\dfrac{1}{2}\left( 3\sqrt{2}-\sqrt{14} \right)}$
Thus, $\tan \dfrac{x}{2}=\dfrac{3\sqrt{2}+\sqrt{14}}{3\sqrt{2}-\sqrt{14}}=\dfrac{{{\left( 3\sqrt{2}+\sqrt{14} \right)}^{2}}}{{{\left( 3\sqrt{2} \right)}^{2}}-{{\left( \sqrt{14} \right)}^{2}}}$
More simplifying, \[\tan \dfrac{x}{2}=\dfrac{{{\left( 3\sqrt{2} \right)}^{2}}+{{\left( \sqrt{14} \right)}^{2}}+2.3\sqrt{2}.\sqrt{14}}{18-14}=\dfrac{18+14+12\sqrt{7}}{4}=\dfrac{32+12\sqrt{7}}{4}=8+3\sqrt{7}\]

So, the correct answer is “Option a”.

Note: In this problem, we have generally used the double angle formula to get our results. There are three forms of double angle formula for cosine. We have used two forms here in the problem, and the one we have not used is $\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x$ . Then, we only need to know one of them as we can derive the other two by the Pythagorean formula.