# Verifying an Identity

• May 29th 2014, 11:04 AM
cdbowman42
Verifying an Identity
tan (x/2 + Π/4) = sec x + tan x

Here is my attempt:

Substituting 1 for tan (Π/4) and applying the tangent of a sum identity gives

[tan (x/2) +1] / [1-tan(x/2)] = 1/cos x + sin x/cos x

Using the half angle identity for tangent, tan(x/2) = sin x /(1 + cos x) and simplifying the complex fraction gives

(sin x + 1 + cos x) / (1 + cos x - sin x) = (1 + sin x) / cos x

• May 29th 2014, 11:10 AM
prasum
Re: Verifying an Identity
u have got the result (1+sin x)/cos x=1/cos x+sin x/cos x=sec x+tan x
• May 29th 2014, 11:36 AM
cdbowman42
Re: Verifying an Identity
(1 + sin x )/cos x was just me changing the right side, sec x + tan x, in terms of cosine and sine. I'm trying to figure out how to get the left side of the equation (sin x + 1 + cos x) / (1 + cos x - sin x) to equal (1 + sin x )/cos x. I cannot find that relationship.
• May 29th 2014, 01:52 PM
Soroban
Re: Verifying an Identity
Hello, cdbowman42!

Quote:

$\text{Prove: }\:\tan\left(\tfrac{x}{2} + \tfrac{\pi}{4}\right) \:=\: \sec x + \tan x$

Never ever play with both sides of an identity.
Choose one side and show that it is equal to the other side.

$LHS \;=\;\tan\left(\tfrac{x}{2} +\tfrac{\pi}{4}\right) \;=\;\frac{\tan\frac{x}{2} + \tan\frac{\pi}{4}}{1 - \tan\frac{x}{2}\tan\frac{\pi}{4}} \;=\;\frac{\tan\frac{x}{2} + 1}{1 - \tan\frac{x}{2}}\;=\; \frac{\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} + 1}{1 - \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}}$

Multiply by $\tfrac{\cos\frac{x}{2}}{\cos\frac{x}{2}}\!;\;\; \frac{\sin\frac{x}{2} + \cos\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}}$

Multlply by $\tfrac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} + \sin\frac{x}{2}}\!:\;\frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} + \sin\frac{x}{2}}\cdot \frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}}$

The numerator is: . $\left(\cos\tfrac{x}{2} + \sin\tfrac{x}{2}\right)^2 \;=\; \cos^2\!\tfrac{x}{2} + 2\sin\tfrac{x}{2}\cos\tfrac{x}{2} + \sin^2\!\tfrac{x}{2} \;=\;1+\sin x$

The denominator is: . $\cos^2\!\tfrac{x}{2} - \sin^2\!\tfrac{x}{2} \;=\;\cos x$

We have: . $\frac{1 + \sin x}{\cos x} \;=\;\frac{1}{\cos x} + \frac{\sin x}{\cos x} \;=\;\sec x + \tan x \;=\;RHS$
• May 29th 2014, 07:44 PM
Prove It
Re: Verifying an Identity
Quote:

Originally Posted by Soroban
Hello, cdbowman42!

Never ever play with both sides of an identity.
Choose one side and show that it is equal to the other side.

$LHS \;=\;\tan\left(\tfrac{x}{2} +\tfrac{\pi}{4}\right) \;=\;\frac{\tan\frac{x}{2} + \tan\frac{\pi}{4}}{1 - \tan\frac{x}{2}\tan\frac{\pi}{4}} \;=\;\frac{\tan\frac{x}{2} + 1}{1 - \tan\frac{x}{2}}\;=\; \frac{\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} + 1}{1 - \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}}$

Multiply by $\tfrac{\cos\frac{x}{2}}{\cos\frac{x}{2}}\!;\;\; \frac{\sin\frac{x}{2} + \cos\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}}$

Multlply by $\tfrac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} + \sin\frac{x}{2}}\!:\;\frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} + \sin\frac{x}{2}}\cdot \frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}}$

The numerator is: . $\left(\cos\tfrac{x}{2} + \sin\tfrac{x}{2}\right)^2 \;=\; \cos^2\!\tfrac{x}{2} + 2\sin\tfrac{x}{2}\cos\tfrac{x}{2} + \sin^2\!\tfrac{x}{2} \;=\;1+\sin x$

The denominator is: . $\cos^2\!\tfrac{x}{2} - \sin^2\!\tfrac{x}{2} \;=\;\cos x$

We have: . $\frac{1 + \sin x}{\cos x} \;=\;\frac{1}{\cos x} + \frac{\sin x}{\cos x} \;=\;\sec x + \tan x \;=\;RHS$

Sorry Soroban, I completely disagree. While it is more elegant to start with one side and go through logical steps to get to the other, it is still perfectly valid to prove the truth of an identity by working with both sides of an equation and showing that they are equivalent to the same thing. The transitive property states that if two things are equal to the same thing, then they are equal to each other.