The Student Room Group

inverse trig functions

Need some help with part 9ii.

I got a to be 1/2pi for part I. But not sure how to do the second bit which may seem a bit daft.

So tan^-1x=1/4pi

How do I get x
Thanks in advance :smile:20141115_213542.jpg
Reply 1
Original post by Super199
Need some help with part 9ii.

I got a to be 1/2pi for part I. But not sure how to do the second bit which may seem a bit daft.

So tan^-1x=1/4pi

How do I get x
Thanks in advance :smile:20141115_213542.jpg


I cannot see the question but you need to "tan" both sides

then x = tan(pi/4)
Reply 2
Original post by TeeEm
I cannot see the question but you need to "tan" both sides

then x = tan(pi/4)

Derp of course.

Can you see the image?


Need help with iv as well.

How do you solve them when you equate the two inverse tans together?
Reply 3
Original post by Super199
Derp of course.

Can you see the image?


Need help with iv as well.

How do you solve them when you equate the two inverse tans together?


f(x) = g(x)

f(x)-g(x)=0

let h(x) = f(x)-g(x)

h(1.535) = ...
h(1.545) = ...

change of sign etc

any good?
Reply 4
Original post by TeeEm
f(x) = g(x)

f(x)-g(x)=0

let h(x) = f(x)-g(x)

h(1.535) = ...
h(1.545) = ...

change of sign etc

any good?

Can you explain your 3rd line. I would never have thought of doing that.
Reply 5
Original post by Super199
Can you explain your 3rd line. I would never have thought of doing that.


these are the bounds of 1.54

well if you achieve a change of sign on these bounds your solution satisfies

1.535 < solution < 1.545

therefore it must round to 1.54

I hope it makes sense
Reply 6
Original post by TeeEm
these are the bounds of 1.54

well if you achieve a change of sign on these bounds your solution satisfies

1.535 < solution < 1.545

therefore it must round to 1.54

I hope it makes sense


I think so. Bit more practice is needed I think.

Thanks :smile:
Reply 7
Original post by Super199
I think so. Bit more practice is needed I think.

Thanks :smile:


naturally

:smile:

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