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