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How cot inverse (x) is (pie - tan inverse (1÷ x) ) and sec inverse x is simply cos inverse (1÷x) when x <0 ?

difficulty advanced
Guest Mar 13, 2015

Best Answer 

 #1
avatar+94101 
+5

 

How cot inverse (x) is (pie - tan inverse (1÷ x) ) and sec inverse x is simply cos inverse (1÷x) when x

Lets see if I can make sense of your question :)

 

How come         $$cot^{-1}(x)\;\; is\;\;\; \pi - tan^{-1}\left(\frac{1}{x}\right)$$

 

and           $$sec^{-1}(x)=cos^{-1}\left(\frac{1}{x}\right) \qquad when\;\; x <0 ?$$

 

 

from my graphs I can see that

 

cos and sec graphs             https://www.desmos.com/calculator/3ekbqgy7ve

Cot and tan graph              https://www.desmos.com/calculator/j1xb33odx7

 

$$\\sec^{-1}x=cos^{-1}\left(\frac{1}{x}\right)\qquad $when $ x\le-1\;\;and\;\; x\ge 1\\\\
cot^{-1}(x)= tan^{-1}\left(\frac{1}{x}\right)\qquad when\;\;x>0\\
and\\
cot^{-1}(x)=\pi\;\textcolor[rgb]{1,0,0}{+}\; tan^{-1}\left(\frac{1}{x}\right)\qquad when\;\;x<0\\\\$$
 

 

I have been mucking around with this and I need to give it more thought.

Perhaps another mathematician would like to jump in here. 

 

 

 

 

How cot inverse (x) is (pie - tan inverse (1÷ x) ) and sec inverse x is simply cos inverse (1÷x) when x
Melody  Mar 14, 2015
 #1
avatar+94101 
+5
Best Answer

 

How cot inverse (x) is (pie - tan inverse (1÷ x) ) and sec inverse x is simply cos inverse (1÷x) when x

Lets see if I can make sense of your question :)

 

How come         $$cot^{-1}(x)\;\; is\;\;\; \pi - tan^{-1}\left(\frac{1}{x}\right)$$

 

and           $$sec^{-1}(x)=cos^{-1}\left(\frac{1}{x}\right) \qquad when\;\; x <0 ?$$

 

 

from my graphs I can see that

 

cos and sec graphs             https://www.desmos.com/calculator/3ekbqgy7ve

Cot and tan graph              https://www.desmos.com/calculator/j1xb33odx7

 

$$\\sec^{-1}x=cos^{-1}\left(\frac{1}{x}\right)\qquad $when $ x\le-1\;\;and\;\; x\ge 1\\\\
cot^{-1}(x)= tan^{-1}\left(\frac{1}{x}\right)\qquad when\;\;x>0\\
and\\
cot^{-1}(x)=\pi\;\textcolor[rgb]{1,0,0}{+}\; tan^{-1}\left(\frac{1}{x}\right)\qquad when\;\;x<0\\\\$$
 

 

I have been mucking around with this and I need to give it more thought.

Perhaps another mathematician would like to jump in here. 

 

 

 

 

How cot inverse (x) is (pie - tan inverse (1÷ x) ) and sec inverse x is simply cos inverse (1÷x) when x
Melody  Mar 14, 2015
 #2
avatar+94101 
0

Please can someone else discuss this one.   :)

I know my CDD is probably causing problems here but I always seem to get to it late at night and it has confused me :(

Melody  Mar 15, 2015
 #3
avatar+890 
0

For the first relationship, cot inverse and tan inverse (to borrow the notation), have different principal ranges for negative values of x. The principal range for the inverse cotangent is pi/2 to pi while for the inverse tangent it is -pi/2 to 0. What that means is that cot inverse will return an angle in the second quadrant while tan inverse returns an angle in the fourth quadrant. The two angles differ by pi, so pi has to be added to the negative angle returned by tan inverse to achieve equality with cot inverse. (So should there be a plus sign rather than a negative sign beteween the two terms on the RHS ?)

(Incidently, not everyone uses the same principal range for cot inverse. For some, including Wolfram I think, the principal range is -pi/2 to 0, the same as for tan inverse. In that case the pi would not be needed.)

 

With the cos inverse and sec inverse though the principal ranges are the same, pi/2 to pi, so no additional pi is needed.

Bertie  Mar 15, 2015

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