The Student Room Group

FP2 trig equation

Hi, the question is: "Find the point of intersection of the graphs y=arctan(x)y=arctan(x) and y=arccos(x) y=arccos(x) . Find the gradient of each graph at the point of intersection. This is from my FP2 textbook.

I've done the following so far: arctan(x)arccos(x)=0 arctan(x)-arccos(x)=0
cos[arctan(x)arccos(x)]=1 cos[arctan(x)-arccos(x)]=1
cos(arctan(x))x+sin(arctan(x))sin(arccos(x))=1 cos(arctan(x))x+sin(arctan(x))sin(arccos(x))=1
x(1+x2)+x(1+x2)(1x2)=1 \frac{x}{\sqrt(1+x^2)}+\frac{x}{\sqrt(1+x^2)}\sqrt(1-x^2)=1

Now I'm stuck.
Original post by maths learner
Hi, the question is: "Find the point of intersection of the graphs y=arctan(x)y=arctan(x) and y=arccos(x) y=arccos(x) . Find the gradient of each graph at the point of intersection. This is from my FP2 textbook.

I've done the following so far: arctan(x)arccos(x)=0 arctan(x)-arccos(x)=0
cos[arctan(x)arccos(x)]=1 cos[arctan(x)-arccos(x)]=1
cos(arctan(x))x+sin(arctan(x))sin(arccos(x))=1 cos(arctan(x))x+sin(arctan(x))sin(arccos(x))=1
x(1+x2)+x(1+x2)(1x2)=1 \frac{x}{\sqrt(1+x^2)}+\frac{x}{\sqrt(1+x^2)}\sqrt(1-x^2)=1

Now I'm stuck.


Have you considered rewriting the equations as x=tanyx = \tan y and x=cosyx=\cos y and solving these?

:smile:
Original post by Mr M
Have you considered rewriting the equations as x=tanyx = \tan y and x=cosyx=\cos y and solving these?

:smile:


Ah I'll try that. Thank you.
Original post by Mr M
Have you considered rewriting the equations as x=tanyx = \tan y and x=cosyx=\cos y and solving these?

:smile:


So I solved it that way and got:

tany=cosy tany=cosy
tanycosy=0 tany-cosy=0
sinycosycosy=0\frac{siny}{cosy}-cosy=0
siny(1sin2y)=0 siny-(1-sin^2y)=0
sin2y+siny1=0 sin^2y+siny-1=0
solving this gave 12+52-\frac{1}{2}+\frac{\sqrt5}{2} .

This isn't the same as the answer in the book, can you see where I've gone wrong? Never mind I didn't inverse sin. Thanks.
(edited 9 years ago)

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