Help me understand trigonometry.

Question

$\frac{\cos \frac{ x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}} = \sec + \tan x$

Using the half angle identity.

$\frac{\sqrt{\frac{1 + \cos x}{2}} + \sqrt{\frac{1 - cos x}{2}}}{\sqrt{\frac{1 + cos x}{2}} - \sqrt{\frac{1 - cos x}{2}}} = \sec x + \tan x$

The denomintaor of the fraction has two radicals in it. I'll rationalize the denominator.

$\frac{\sqrt{\frac{1 + cos x}{2}} + \sqrt{\frac{1 - cos x}{2}}}{\sqrt{\frac{1 + cos x}{2}} - \sqrt{\frac{ 1 - cos x}{2}}} \cdot \frac{\sqrt{\frac{1 + cos x}{2}} + \sqrt{\frac{1 - cos x}{2}}}{\sqrt{\frac{1 + cos x}{2}} + \sqrt{\frac{ 1 - cos x}{2}}}$ Q. Why are the left and right side different? Could someone break-down the numerator's multiplication, step-by-step? I'm having trouble understanding the steps; for example, where the $2$ and the $\cos^2$ and the $4$ in the numerator, from? Why does the $-$ sign make such a change in the denominator, between the two square-rooted fractions on the left side? $\frac{\frac{1 + cos x}{2} + {2}\sqrt{\frac{1 - \cos^2 x}{4}} + \frac{1 - cos x}{2}}{\frac{1 + cos x}{2} - \frac{1 - cos x}{2}}$

$\frac{\frac{1 + cos x + 1 - cos x}{2} + {2}\sqrt{\frac{1 - cos^2 x}{2}}}{\frac{1 + cos x - 1 + cos x}{2}}$

$\frac{1 + \sqrt{1 - \cos^2 x}}{\cos x}$ Q. How does the $2$ dissappear before the square-root symbol?

Answer 1

I think , you want to show $$\frac{cos(\frac{x}{2})+sin(\frac{x}{2})}{cos(\frac{x}{2})-sin(\frac{x}{2})}=\\ \frac{cos(\frac{x}{2})+sin(\frac{x}{2})}{cos(\frac{x}{2})-sin(\frac{x}{2})}\times \frac{cos(\frac{x}{2})+sin(\frac{x}{2})}{cos(\frac{x}{2})+sin(\frac{x}{2})}=\\ \frac{(cos(\frac{x}{2})+sin(\frac{x}{2}))^2}{cos^2(\frac{x}{2})-sin^2(\frac{x}{2})}=\\ \frac{cos^2(\frac{x}{2})+sin^2(\frac{x}{2})+2cos(\frac{x}{2}).sin(\frac{x}{2})}{cos (\frac{2x}{2})}=\\ \frac{1+2cos(\frac{x}{2}).sin(\frac{x}{2})}{cos (\frac{2x}{2})}=\\ \frac{1+sin(\frac{2x}{2})}{cos (\frac{2x}{2})}=\\ \frac{1+sin x}{cos x}=\\ \frac{1}{cos x}+\frac{sin x}{cos x}=\\sec x +tan x$$ by your way : $$\large \frac{\sqrt{\frac{1 + \cos x}{2}} + \sqrt{\frac{1 - cos x}{2}}}{\sqrt{\frac{1 + cos x}{2}} - \sqrt{\frac{1 - cos x}{2}}} \times \large \frac{\sqrt{\frac{1 + \cos x}{2}} + \sqrt{\frac{1 - cos x}{2}}}{\sqrt{\frac{1 + cos x}{2}} + \sqrt{\frac{1 - cos x}{2}}}\\ \large \frac{\frac{1 + \cos x}{2} + \frac{1 - \cos x}{2}+2\sqrt{\frac{1 - cos^2 x}{4}}}{\frac{1 + cos x}{2} - \frac{1 - cos x}{2}}=\\ \large \frac{1+2\sqrt{\frac{1 - cos^2 x}{4}}}{cosx}=\\ \frac{1+2\frac{\sqrt{1 - cos^2 x}}{\sqrt 4}}{cosx}=\\ \frac{1+2\frac{\sqrt{1 - cos^2 x}}{2}}{cosx}=\\ \frac{1+\sqrt{1 - cos^2 x}}{cosx}=\\\frac{1+\sqrt{sin^2 x}}{cosx}=\\\frac{1+sinx}{cosx}=\\= \sec x + \tan x$$

Answer 2

The main problem with your argument is trying to use the half-angle formulas: it is false that $$ \cos\frac{x}{2}=\sqrt{\frac{1+\cos x}{2}} $$ If you add $\pm$ in front of the square root, you just make things worse, because this means that only one of the signs is right. And if $-$ is good for the cosine, it might not be for the sine. So you'd need several different cases. And the formulas get more and more complicated.

Much better is to set $y=x/2$ and start from $$ \frac{\cos y+\sin y}{\cos y-\sin y} $$ but taking into account that if you used the half-angle formulas, you surely would rationalize: $$ \frac{\cos y+\sin y}{\cos y-\sin y}= \frac{(\cos y+\sin y)^2}{\cos^2y-\sin^2y} $$ Now $$ (\cos y+\sin y)^2=\cos^2y+2\cos y\sin y+\sin^2y=1+2\cos y\sin y=1+\sin2y $$ and $$ \cos^2y-\sin^2y=\cos2y $$ so we end up with $$ \frac{\cos y+\sin y}{\cos y-\sin y}= \frac{1+\sin2y}{\cos2y}=\frac{1+\sin x}{\cos x}=\sec x+\tan x $$

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The second part of the question is completely unrelated to the first one, and it is the object of a previous question of yours: How to create an identity for $\sin \frac{x}{4}$

Answer 3

What is $\frac{\sqrt{\frac{1 + cos x}{2}} + \sqrt{\frac{1 - cos x}{2}}}{\sqrt{\frac{1 + cos x}{2}} - \sqrt{\frac{ 1 - cos x}{2}}} \cdot \frac{\sqrt{\frac{1 + cos x}{2}} + \sqrt{\frac{1 - cos x}{2}}}{\sqrt{\frac{1 + cos x}{2}} + \sqrt{\frac{ 1 - cos x}{2}}}$?

Let $a = \sqrt{\frac{1 + cos x}{2}}$ and $b = \sqrt{\frac{1 - cos x}{2}}$

Then $\frac{\sqrt{\frac{1 + cos x}{2}} + \sqrt{\frac{1 - cos x}{2}}}{\sqrt{\frac{1 + cos x}{2}} - \sqrt{\frac{ 1 - cos x}{2}}} \cdot \frac{\sqrt{\frac{1 + cos x}{2}} + \sqrt{\frac{1 - cos x}{2}}}{\sqrt{\frac{1 + cos x}{2}} + \sqrt{\frac{ 1 - cos x}{2}}}=\frac{a+b}{a-b}\frac{a+b}{a+b} =\frac {(a+b)^2}{(a-b)(a+b)} =\frac {a^2 + 2ab + b^2}{a^2 - b^2}$.

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

$b^2 = \frac{1 - \cos x}2$

$a^2 + b^2 = \frac{1 + \cos x+1 - \cos x}2= \frac 22 = 1$

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

$2ab = 2\sqrt{\frac{(1+\cos x)(1-\cos x)}{2*2}} = 2\sqrt{\frac{(1 - \cos^2 x)}{4}}= 2\frac {\sqrt{1-\cos^2 x}}2 = \sqrt{1- \cos^2 x}$

So $\frac{\sqrt{\frac{1 + cos x}{2}} + \sqrt{\frac{1 - cos x}{2}}}{\sqrt{\frac{1 + cos x}{2}} - \sqrt{\frac{ 1 - cos x}{2}}} \cdot \frac{\sqrt{\frac{1 + cos x}{2}} + \sqrt{\frac{1 - cos x}{2}}}{\sqrt{\frac{1 + cos x}{2}} + \sqrt{\frac{ 1 - cos x}{2}}}=\frac{a+b}{a-b}\frac{a+b}{a+b} =\frac {(a+b)^2}{(a-b)(a+b)} =\frac {a^2 + 2ab + b^2}{a^2 - b^2}=$

$\frac {1 + \sqrt{1- \cos^2 x}}{\cos x}$