arcsin/cos/tan Watch

awargi
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can somebody explain to me wat does "arcsin" "arccos" "arctan" mean ???


thnx alot.




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MonkeyMajik
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it's just sin^-1, cos^-1, tan^-1
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miml
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They are the inverse trig functions.

So the inverse of sin is arcsin, etc. I'm not 100% sure exactly what you want to know about them.
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eulerwaswrong
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(Original post by awargi)
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can somebody explain to me wat does "arcsin" "arccos" "arctan" mean ???


thnx alot.




--------------------------------
well i suppose you are familiar with sin, cos and tan. Its just the inverse function of those. What level are you at GCSE i assume?

Often written as sin^-1 etc
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fishpie57
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Not to be confused with cosec x, sec x and cot x... Which are 1/sin x, 1/cos x and 1/tan x respectively.

When you have f(x), its inverse function [f^-1(x)] is f(x) reflected in the line y=x. Arcsin, arccos and arctan are the inverses you get on the calculator, where it says sin^-1, like everyone else has said.

However, in this case ^-1 doesn't mean 'one over...', it's just a way of writing the inverse function. I know this bit confused me when my maths teacher tried to explain the difference - hope it's of some use!
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struz
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sorry for butting in but i was about to post a thread relating to this topic but since there's already one, how are \frac{1}{sin x} and arcsin different? when would you use arcsin instead of \frac{1}{sin x}?

and also here be the question,
(b) Given that x = 4 sin (2y + 6), find \frac{dy}{dx} in terms of x.

Managed to get this:

\frac{dy}{dx} = \frac{1}{8 cos (2y +6)} but then it gives as the final answer \frac{1}{8 cos (arcsin \frac{x}{4})} ...which just made me ridiculously confused.

could anyone please explain how they get there? ta for any help
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nuodai
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(Original post by struz)
sorry for butting in but i was about to post a thread relating to this topic but since there's already one, how are \frac{1}{sin x} and arcsin different? when would you use arcsin instead of \frac{1}{sin x}?

and also here be the question,
(b) Given that x = 4 sin (2y + 6), find \frac{dy}{dx} in terms of x.

Managed to get this:

\frac{dy}{dx} = \frac{1}{8 cos (2y +6)} but then it gives as the final answer \frac{1}{8 cos (arcsin \frac{x}{4})} ...which just made me ridiculously confused.

could anyone please explain how they get there? ta for any help
You've got it in terms of y; you need to rearrange x = 4\sin (2y + 6) to get y in terms of x.
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fishpie57
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(Original post by struz)
sorry for butting in but i was about to post a thread relating to this topic but since there's already one, how are \frac{1}{sin x} and arcsin different? when would you use arcsin instead of \frac{1}{sin x}?

and also here be the question,
(b) Given that x = 4 sin (2y + 6), find \frac{dy}{dx} in terms of x.

Managed to get this:

\frac{dy}{dx} = \frac{1}{8 cos (2y +6)} but then it gives as the final answer \frac{1}{8 cos (arcsin \frac{x}{4})} ...which just made me ridiculously confused.

could anyone please explain how they get there? ta for any help
Essentially, arcsin, arccos etc. are the buttons you press on the calculator for solving something like sin x = 0.5, so you do arcsin [sin^-1] and get x=30. cosec x is 1/sin x, and I think you'd tend to use this one in trying to solve other equations where you might need a trig identity.
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fishpie57
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(Original post by struz)
sorry for butting in but i was about to post a thread relating to this topic but since there's already one, how are \frac{1}{sin x} and arcsin different? when would you use arcsin instead of \frac{1}{sin x}?

and also here be the question,
(b) Given that x = 4 sin (2y + 6), find \frac{dy}{dx} in terms of x.

Managed to get this:

\frac{dy}{dx} = \frac{1}{8 cos (2y +6)} but then it gives as the final answer \frac{1}{8 cos (arcsin \frac{x}{4})} ...which just made me ridiculously confused.

could anyone please explain how they get there? ta for any help
If x=4sin(2y+6), rearrange this to get an expression in terms of x for 8cos(2y+6) and substitute it in. Here goes:

x^2=16sin^2(2y+6)

= 16(1-cos^2(2y+6))

= 16 - 16cos^2(2y+6)

16cos^2(2y+6) = 16 - x^2

4cos(2y+6) = (16-x^2)^1/2

8cos(2y+6) = 2(16-x^2)^1/2

Then substitute 8cos(2y+6) for the expression in terms of x and simplify!

Hope that's right... :eek:
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struz
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(Original post by nuodai)
You've got it in terms of y; you need to rearrange x = 4\sin (2y + 6) to get y in terms of x.
ok this attempt is no doubt going to be horrific but is here goes...

\frac{x}{4sin} = (2y + 6) (i don't think that's actually mathematically allowed/possible?!)

and then divide by 2 and minus 6? ha something went wrong...sorry, i still can't work it out!
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fishpie57
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(Original post by struz)
ok this attempt is no doubt going to be horrific but is here goes...

\frac{x}{4sin} = (2y + 6) (i don't think that's actually mathematically allowed/possible?!)

and then divide by 2 and minus 6? ha something went wrong...sorry, i still can't work it out!
No no no - you can't split up sin(2y+6)!!! There's no such thing as 4sin!! :eek3:
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struz
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(Original post by fishpie57)
No no no - you can't split up sin(2y+6)!!! There's no such thing as 4sin!! :eek3:
hahaha :o: oh the shame, i'm sorry, i'm just a bit **** at grasping most mathematical concepts!

thanks for your post above, it is the right answer! but the stage just before it in the mark scheme gives \frac{1}{8 cos (arcsin \frac{x}{4})} and i'm just a bit stumped as to how to get there?
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nuodai
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(Original post by struz)
ok this attempt is no doubt going to be horrific but is here goes...

\frac{x}{4sin} = (2y + 6) (i don't think that's actually mathematically allowed/possible?!)

and then divide by 2 and minus 6? ha something went wrong...sorry, i still can't work it out!
You should know that the inverse of \sin x is \sin^{-1} x... that is, A = \sin B \Rightarrow B = \sin^{-1} A. Apply this here:

\newline x = 4\sin(2y + 6)\newline

\Rightarrow \dfrac{x}{4} = \sin(2y + 6)\newline

\Rightarrow \sin^{-1} \left( \dfrac{x}{4} \right) = 2y + 6\ (*) \newline

\Rightarrow \sin^{-1} \left( \dfrac{x}{4} \right) - 6 = 2y \newline

\Rightarrow y = \dfrac{1}{2} \Bigg( \sin^{-1} \left( \dfrac{x}{4} \right) - 6 \Bigg)

I put a (*) next to the one that you need, because you can just replace 2y+6 with  \sin^{-1} \left( \dfrac{x}{4} \right) in your answer (which gives you the same as what the actual answer is).
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awargi
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so basically wat u people are trying to tell me is this that

sin^-1 is arcsin and not \frac{1}{sin}
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awargi
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(Original post by struz)
sorry for butting in but i was about to post a thread relating to this topic but since there's already one, how are \frac{1}{sin x} and arcsin different? when would you use arcsin instead of \frac{1}{sin x}?

and also here be the question,
(b) Given that x = 4 sin (2y + 6), find \frac{dy}{dx} in terms of x.

Managed to get this:

\frac{dy}{dx} = \frac{1}{8 cos (2y +6)} but then it gives as the final answer \frac{1}{8 cos (arcsin \frac{x}{4})} ...which just made me ridiculously confused.

could anyone please explain how they get there? ta for any help


is this C3 or C4???
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Daniel Freedman
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(Original post by awargi)
so basically wat u people are trying to tell me is this that

sin^-1 is arcsin and not \frac{1}{sin}
Yes. The notation is a bit confusing, but  \sin^{-1}(x) means the inverse sine function.  (\sin{x})^{-1} would mean  \frac{1}{\sin{x}}
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struz
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(Original post by nuodai)
You should know that the inverse of \sin x is \sin^{-1} x... that is, A = \sin B \Rightarrow B = \sin^{-1} A. Apply this here:

\newline x = 4\sin(2y + 6)\newline

\Rightarrow \dfrac{x}{4} = \sin(2y + 6)\newline

\Rightarrow \sin^{-1} \left( \dfrac{x}{4} \right) = 2y + 6\ (*) \newline

\Rightarrow \sin^{-1} \left( \dfrac{x}{4} \right) - 6 = 2y \newline

\Rightarrow y = \dfrac{1}{2} \Bigg( \sin^{-1} \left( \dfrac{x}{4} \right) - 6 \Bigg)

I put a (*) next to the one that you need, because you can just replace 2y+6 with  \sin^{-1} \left( \dfrac{x}{4} \right) in your answer (which gives you the same as what the actual answer is).
Thank you!

awargi - tis C3.
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