# Trig identities

• Jun 22nd 2009, 03:50 AM
bobred
Trig identities
Hi

Is there any way of converting the function below into one that doesn't involve a quotient?

$\frac{2\sec x}{2+\tan x}$

Thanks
• Jun 22nd 2009, 05:34 AM
mr fantastic
Quote:

Originally Posted by bobred
Hi

Is there any way of converting the function below into one that doesn't involve a quotient?

$\frac{2\sec x}{2+\tan x}$

Thanks

The expression simplifies to $\frac{2}{2 \cos x + \sin x}$ so I don't see a way of avoiding a quotient ....
• Jun 22nd 2009, 05:44 AM
Soroban
Hello, bobred!

Quote:

Is there any way of converting this function
. . into one that doesn't involve a quotient? . . $\frac{2\sec x}{2+\tan x}$

Yes, there is.
But the method may not be within the scope of your course.

We have: . $\frac{\dfrac{2}{\cos x}}{2 + \dfrac{\sin x}{\cos x}}$

Multiply by $\frac{\cos x}{\cos x}\!:\quad\frac{\cos x\left(\dfrac{2}{\cos x}\right)} {\cos x\left(2 + \dfrac{\sin x}{\cos x}\right)} \;=\;\frac{2}{2\cos x + \sin x}$ .[1]

Consider the denominator: . $2\cos x + \sin x$

. . Multiply by $\tfrac{\sqrt{5}}{\sqrt{5}}\!:\quad \tfrac{\sqrt{5}}{\sqrt{5}}(2\cos x + \sin x) \;=\;\sqrt{5}\left(\tfrac{2}{\sqrt{5}}\cos x + \tfrac{1}{\sqrt{5}}\sin x\right)$

. . Let $\theta$ be an angle sucn that: . $\sin\theta = \tfrac{2}{\sqrt{5}},\;\cos\theta = \tfrac{1}{\sqrt{5}}$

. . Then we have: . $\sqrt{5}\left(\sin\theta\cos x + \cos\theta\sin x\right) \;=\;\sqrt{5}\sin(x + \theta)$

Hence, [1] becomes: . $\frac{2}{\sqrt{5}\sin(x + \theta)} \;=\;\frac{2}{\sqrt{5}}\csc(x + \theta)$

Therefore: . $\frac{2\sec x}{2 + \tan x} \;=\;\frac{2}{\sqrt{5}}\csc\bigg[x + \arccos\left(\tfrac{1}{\sqrt{5}}\right)\bigg]$

• Jun 22nd 2009, 06:08 AM
bobred
Now I see.

I had have been given some answers similar but none that have shown how they have arrived at the answer, Thanks
• Jun 22nd 2009, 06:24 AM
bobred
Just one more thing, why multipy by $\sqrt{5}$ and not something else?
• Jun 22nd 2009, 06:49 AM
yeongil
Quote:

Originally Posted by bobred
Just one more thing, why multipy by $\sqrt{5}$ and not something else?

Recall what Soroban wrote:
Quote:

Let $\theta$ be an angle such that: . $\sin\theta = \tfrac{2}{\sqrt{5}},\;\cos\theta = \tfrac{1}{\sqrt{5}}$
You have a right triangle with angle $\theta$ where the opposite side is 2 and the adjacent side is 1. By the Pythagorean Theorem, the hypotenuse has to be $\sqrt{2^1 + 1^2} = \sqrt{5}$.

01
• Jun 22nd 2009, 06:58 AM
bobred
Thanks.

Thought it would be something obvious.