Question

sin x = 1/4, x lies in 2nd quadrant. Find sin x/2, cos x/2, tan x/2.

Answer 1

check the attachment

Answer 2

Answer:

$sin{\frac{x}{2}=\sqrt{\frac{4+\sqrt{15}}{8}}$

$cos{\frac{x}{2}}=\sqrt{\frac{4-\sqrt{15}}{8}}$

$tan{\frac{x}{2}=\sqrt{\frac{4+\sqrt{15}}{4-\sqrt{15}}$

Step-by-step explanation:

Given $sinx=\frac{1}{4}$, x lies in second quadrant.

we have to find the value of $sin{\frac{x}{2}}, cos{\frac{x}{2}}, tan{\frac{x}{2}}$

Since, $sin^2{x}+cos^2{x}=1$

⇒ $cos{x}=\pm{\sqrt{1-sin^2{x}}}$

⇒ $cos{x}=\pm{\sqrt{1-\frac{1}{16}}}=\pm{\frac{\sqrt{15}}{4}}$

Since, x lies in second quadrant

⇒ $cos{x}=-{\frac{\sqrt{15}}{4}}$

Now, by using identity of cos 2x

$2cos^2(\frac{x}{2})=1+cos{x}$

$cos{\frac{x}{2}}=\sqrt{\frac{1}{2}(1+cos{x})}=\sqrt{\frac{1}{2}(1-{\frac{\sqrt{15}}{4}})}=\sqrt{\frac{4-\sqrt{15}}{8}}$

Also,  $sin^2{\frac{x}{2}}=1-cos^2{\frac{x}{2}}$

⇒ $sin{\frac{x}{2}=\sqrt{1-cos^2({\frac{x}{2})}}=\sqrt{1-\frac{4-\sqrt{15}}{8}}=\sqrt{\frac{4+\sqrt{15}}{8}}$

Now,

$tan{\frac{x}{2}=\frac{sin\frac{x}{2}}{cos{\frac{x}{2}}}=\frac{\sqrt{\frac{4+\sqrt{15}}{8}}}{\sqrt{\frac{4-\sqrt{15}}{8}}}=\sqrt{\frac{4+\sqrt{15}}{4-\sqrt{15}}$