simplifying inverse trig functions

[lukelowe]

I really don't know where to go with this one and I haven't been able to find any examples to help.
I need to simplify cot[Cos-1 7/x]. The x variable is what's got me confused.

 

[Deleted member 4993]

I really don't know where to go with this one and I haven't been able to find any examples to help.
I need to simplify cot[Cos-1 7/x]. The x variable is what's got me confused.

I assume - your expression is:

\(\displaystyle cot\left[ cos^{-1}\left(\frac{7}{x}\right)\right]\)

assume:

\(\displaystyle \theta \ = \ cos^{-1}\left(\frac{7}{x}\right)\)

then

\(\displaystyle cos(\theta) \ = \ \frac{7}{x}\)

then

\(\displaystyle sin(\theta) \ = \ \pm\sqrt{1-\left(\frac{7}{x}\right)^2}\)

then

\(\displaystyle cot(\theta) \ = \ ???\)

 

[lukelowe]

So would that make Cot = 7/x over radical 1- 7/x squared or should that be simplified more?

 

[Deleted member 4993]

So would that make Cot = 7/x over radical 1- 7/x squared or should that be simplified more?

That depends on your instructor's choice.

My personal choice would be to simplify further.

 

[christinojacob]

here is the solution

I really don't know where to go with this one and I haven't been able to find any examples to help.
I need to simplify cot[Cos-1 7/x]. The x variable is what's got me confused.

hey write cos-1 7/x as cot-1 7/x²-49

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so ur answer will be 7/x²​-49

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[soroban]

Hello, lukelowe!

I will assume that the angle is in Quadrant 1.

\(\displaystyle \text{Simplify: }\:\cot\left(\cos^{-1}\dfrac{7}{x}\right)\)

Let \(\displaystyle \theta \,=\,\cos^{-1}\dfrac{7}{x}\)
. . Then: .\(\displaystyle \cos\theta \,=\,\dfrac{7}{x} \,=\,\dfrac{adj}{hyp}\)

\(\displaystyle \theta\) is in a right triangle with \(\displaystyle adj = 7,\:hyp = x\)

                  *
           x   *  *
            *     *
         * θ      *
      *  *  *  *  *
            7

Pythagorus: .\(\displaystyle opp \,=\,\sqrt{x^2-49}\)

Therefore: .\(\displaystyle \cot\theta \:=\:\dfrac{adj}{opp} \:=\:\dfrac{7}{\sqrt{x^2-49}} \)