What is the Half Angle Formula for Trigonometric Identities?

[whkoh]

Prove that
$$\sec x + \tan x = \tan \left (\frac{\pi}{4} + \frac{x}{2}\right )$$

I've got to
$$\sec x + \tan x = \frac{1+\sin x}{\cos x}$$
and then I was stuck. Tried half angle but it didn't seem to work.

Help please.

 

[Hurkyl]

There are two sides to the equation -- it sounds like you've only fiddled with the left hand side.

 

[arildno]

Neat!
I've never seen that trig. identity before..

 

[whkoh]

Well, manipulating RHS gives
$$\tan \left (\frac{\pi}{4} + \frac{x}{2} \right )$$
$$= \frac{\tan\frac{\pi}{4}+\tan\frac{x}{2}}{1-\tan\frac{\pi}{4}\tan\frac{x}{2}}$$
$$=\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}$$

and applying half angle to LHS gives
$$\frac{1+\sin x}{\cos x}$$
$$=\frac{1+2\sin\frac{x}{2}\cos\frac{x}{2}}{1-2\sin\frac{x}{2}\sin\frac{x}{2}}$$

Hmm.. how can $$\tan\frac{x}{2}$$ be equal to
$$2\sin\frac{x}{2}\cos\frac{x}{2}}$$
and
$$2\sin\frac{x}{2}\sin\frac{x}{2}}$$ at the same time?

Any help please..

 

[Curious3141]

Here's help. :)

Try to manipulate the more complicated side to get the less complicated side. In this case, work on the RHS to get the LHS. Try not to work from both sides at once.

Let the respective sin, cos and tan trig ratios of x/2 be s, c and t. Let those of x be S, C and T. I'm doing this because I'm really fed up of clunky LaTex.

Taking it from where you left off,

RHS :

$$\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}$$
$$=\frac{(1 + t)^2}{(1-t)(1 + t)}$$
$$=\frac{1 + t^2 + 2t}{1 - t^2}$$
$$=\frac{1 + t^2 + 2t}{\frac{c^2 - s^2}{c^2}}$$
$$=\frac{(1 + t^2 + 2t)(c^2)}{C}$$
$$=\frac{c^2 + s^2 + 2sc}{C}$$
$$=\frac{1 + S}{C}$$
$$=\sec{x} + \tan{x}$$ (QED)

 

[Curious3141]

BTW, the proof (and the original identity) fail for t = -1. In the proof, it's because I multiply the RHS by (1+t)/(1+t). In the orig. identity, the LHS becomes undefined while the RHS remains finite (so the failure is consistent).

 

[whkoh]

Thanks for your help

 

[dextercioby]

You know

$$\sec x+\tan x=\frac{1}{\cos x}+\frac{\sin x}{\cos x}=\frac{1+\sin x}{\cos x}$$ (1)

Th RHS is

$$\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}=\frac{1+\frac{\sin^{2}\frac{x}{2}}{\cos^{2}\frac{x}{2}}+2\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}}{1-\frac{\sin^{2}\frac{x}{2}}{\cos^{2}\frac{x}{2}}}$$

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

(Q.e.d.)

,pretty simple,right...?

Daniel.

 

[arildno]

I think he mentioned that identity in post 1, Daniel..

 

[dextercioby]

Hehe,i thought he went backwards starting with the RHS.

Daniel.