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Math Help - Trig identities

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by bobred View Post
    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 ....
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  3. #3
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    Hello, bobred!

    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]

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  4. #4
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    Now I see.

    I had have been given some answers similar but none that have shown how they have arrived at the answer, Thanks
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  5. #5
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    Just one more thing, why multipy by \sqrt{5} and not something else?
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  6. #6
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    Quote Originally Posted by bobred View Post
    Just one more thing, why multipy by \sqrt{5} and not something else?
    Recall what Soroban wrote:
    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
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  7. #7
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    Thanks.

    Thought it would be something obvious.
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