Arcsin, Arccos, Arctan - please help :)

Hi everybody :smile:

Could somebody be really really nice and perhaps outline what exactly we are going to need to know for edexcel syllabus C3 in terms of Arcsin, Arccos, Arctan?

I just checked the syllabus and it says 'knowledge of Arcsin, Arccos, Arctan and their relationship to sin cos and tan' - which is simple, but then in the book it doesn't explain it very well, and other than knowing they're a reflection of the sin cos tan in y=x I can't quite figure it out.

Would somebody mind explaining it to me please or linking me to a site that does explain it well??

Also could someone tell me the PV equations for working out the next angle for sin cos and tan please?

Thanks so much to anyone who can be arsed to read this, and even more thanks for any replies :biggrin:

Ourkid

Reply 1

Kolya

17

Well:
y = arcsinx
x = siny

That explains why they are a reflection of the normal graphs in y=x (ie. they are inverse functions)

I'm not sure there is much more to it than that. Make sure you look at past papers to see what types of questions have come up before, and make sure you are prepared for them! :smile:

(I know a question came up in the Jan 07 C3)

Reply 2

Chewwy

17

all you really need to know...

arcsin, arccos, arctan, are the inverses of sin, cos, tan respectively (hence all the reflections in y=x stuff).

in order for them to be functions, we must restrict the range of the initial functions so that they are 1-to-1, hence the inverses are 1-to-1 too.

if y = f(x), x = f^-1(y)

Reply 3

Ourkid

OP

12

Wow and that's it?

In the book they start going into detail about drawing the graph and then finding things like the value of arccos 0, and arctan (-1/root3)... which has left me thinking wtf?

Thanks guys - much appreciated :smile:

Ourkid

Reply 4

Kolya

17

Ourkid

Wow and that's it?

In the book they start going into detail about drawing the graph and then finding things like the value of arccos 0, and arctan (-1/root3)... which has left me thinking wtf?
Ourkid

The theory above allows you to solve those problems.

For drawing the graph, just do a faint pencil sketch of the normal trig function and y=x. Then draw in the reflection!

For finding the values,
y = arccos 0
cos y = 0
Then look at the graph of cos y and cheerido! cos y = 0 has solutions at pi/2, 3pi/2 :smile:

The theory is not difficult, but because the inverse trig functions are on the fringes of the syllabus, when they do come up in the exam, many people do badly on them.

Reply 5

Ourkid

OP

12

Ahh I see.

Thanks everybody.

Rather than me clogging up the forum with another thread I hope you don't mind but I'm going to ask a few trig questions that have been bothering me.

I just answered the question:

Solve tanx = 2cotx in -180 =< x =< 90

So I've got:

tan x = 2/tanx

tan^2x = 2

tanx = +- root2

x = 54.7 - which is my principal value

So the only other value I can have is:

pv + 180n (n=1) ---> 54.7 + 180 = 234.7

This isn't in the answer so I've tried the cast diagram.

There is 54.7 in the A quadrant, then the opposite angle 180 degrees across... which is 234.7

The answer says it should be +- 54.7 and -125.3

Now I can't understand why.

I know that -125.3 is just the 360 - 234.7 but I don't see why it's going clockwise now when usually you draw it in anti clockwise

Am I doing something wrong here?

Thanks again everybody :biggrin:

Ourkid

Reply 6

Kolya

17

Ourkid

Ahh I see.

Thanks everybody.

Rather than me clogging up the forum with another thread I hope you don't mind but I'm going to ask a few trig questions that have been bothering me.

I just answered the question:

Solve tanx = 2cotx in -180 =< x =< 90

So I've got:

tan x = 2/tanx

tan^2x = 2

tanx = +- root2

tanx = +- root2

You may want to completely ignore me, but I like to solve these questions with the aid of a sketch.

Okay, so tan x = +- rt 2. Draw the graph of tan x in your range given, and then draw lines across the page at + rt. 2 (~1.4), and at -rt. 2 (~ -1.4). Then you can type in x = arctan (rt. 2) on your calculator, and get the pv of 54.7. Look at your graph and find where that x=54.7 value is. Now look at where else y=rt. 2 crosses your tanx line. It crosses it a little before x=0, and knowing the shape of tanx, you can deduce that value is x=-54.7. It also crosses the tanx graph a little after -180, and above the x axis. Knowing the shape of the tan graph gives this value as 54.7 - 180 = -125.3. This is because tanx cycles around every 180 degrees. So x = -125.3, -54.7, 54.7. Done. :smile:

Reply 7

Ourkid

OP

12

Oooh I like it very mooch.

Thanks so much :biggrin:

Ourkid

Arcsin, Arccos, Arctan - please help :)

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