IF sin x + cos x = root over 7/2 where x belongs to A , then tan x/2 = - Maths - Trigonometric Functions

Question

IF sin x + cos x = root over 7/2 where x belongs to A , then tan
x/2 =?

Aakash Sharma · Accepted Answer

Given:

$\sin x + \cos x = \frac{\sqrt{7}}{2} ........ (1) \Rightarrow {(\sin x + \cos x)}^{2} = (\frac{\sqrt{7}}{2})^{2} \Rightarrow \sin^{2} x + \cos^{2} x + 2 \sin x \cos x = \frac{7}{4} \Rightarrow 1 + 2 \sin x \cos x = \frac{7}{4} \Rightarrow 2 \sin x \cos x = \frac{7}{4} - 1 = \frac{3}{4} ....... (2)$

Now,

${(\cos x - \sin x)}^{2} = \cos^{2} x + \sin^{2} x - 2 \sin x \cos x \Rightarrow \cos x - \sin x = \frac{1}{\sqrt{2}} ...... (3)$

Adding (1) and (2) we get

$2 \cos x = \frac{\sqrt{7}}{2} + \frac{1}{\sqrt{2}} = \frac{\sqrt{7} + \sqrt{2}}{2} \Rightarrow \cos x = \frac{\sqrt{7} + \sqrt{2}}{2} also, \tan \frac{x}{2} = \sqrt{\frac{1 - \cos x}{1 + \cos x}} = \sqrt{\frac{1 - \frac{\sqrt{7} + \sqrt{2}}{2}}{1 + \frac{\sqrt{7} + \sqrt{2}}{2}}} = \sqrt{\frac{(2 - (\sqrt{7} + \sqrt{2})) (2 - \sqrt{7} + \sqrt{2})}{(2 + (\sqrt{7} + \sqrt{2})) (2 - \sqrt{7} + \sqrt{2})}} = \sqrt{\frac{4 + 7 + 2 + 2 \sqrt{14} - 4 \sqrt{7} - 4 \sqrt{2}}{4 - 7 - 2 - 2 \sqrt{14}}} = \sqrt{\frac{13 + 2 \sqrt{14} - 4 \sqrt{7} - 4 \sqrt{2}}{- 5 - 2 \sqrt{14}}}$

IF sin x + cos x = root over 7/2 where x belongs to A , then tan x/2 =?