Hi

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

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

Thanks

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- Jun 22nd 2009, 03:50 AMbobredTrig identities
Hi

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

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

Thanks - Jun 22nd 2009, 05:34 AMmr fantastic
- Jun 22nd 2009, 05:44 AMSoroban
Hello, bobred!

Quote:

Is there any way of converting this function

. . into one that doesn't involve a quotient? . . $\displaystyle \frac{2\sec x}{2+\tan x}$

But the method may not be within the scope of your course.

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

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

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

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

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

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

Therefore: .$\displaystyle \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 AMbobred
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 AMbobred
Just one more thing, why multipy by $\displaystyle \sqrt{5}$ and not something else?

- Jun 22nd 2009, 06:49 AMyeongil
Recall what Soroban wrote:

Quote:

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

01 - Jun 22nd 2009, 06:58 AMbobred
Thanks.

Thought it would be something obvious.