Circular functions

[Hielly]

Hey how would i do this question

[NE2000]

Sort of a giveaway hint:

[Hielly]

hey thanks.

why is it sin pi/2 - x ?, it's saying cos x is the same as sin (pi/2 - x) but what quadrant is this?

[NE2000]

It doesn't actually matter.

Using degrees:

cos(60) = 0.5 = sin(90 - 60) = sin(30) = 0.5
cos(120) = -0.5 = sin(90 - 120) = sin(-30) = -sin(30) = -0.5
cos(240) = -0.5 = sin(90 - 240) = sin(-150) = -sin(150) = -sin(30) = -0.5
cos(300) = 0.5 = sin(90 - 300) = sin(-210) = -sin(210) = sin(30) = 0.5

As you can see, x can be in any quadrant and it still works

[Hielly]

okay thanks
also how woull i find the x intercepts of y=2cos(3x-pi/4)
im stuck on the bit where cos(3x-pi/4)=0, so which quadrant for pos or neg is it in?neither?

[TrueTears]

Let

Now solve for

The result should be quite obvious now.

[Hielly]

hrm wait, i dont quite get it. maybe i should have typed out the whole question.

sketch the graphs of the following for [-2pi,2pi]

so
cos(3x-pi/4)=0
3x-pi/4=pi/2

not sure what step next.. im using the essentials methods to do this question


thanks

[TrueTears]

When is equal to ?

Every where .

If you can understand this and imagine the unit circle, the rest should be fine.

[Hielly]

no.. i don't recall learning that :S. When i look at the essentials explanation they didn't mention what you just said.. do you know whereabouts they explain this is the essentials?

[TrueTears]

This is where Essentials fails.

I suggest before doing these questions you need to understand where these things come from. A good grasps of the concepts will lead you to the deviation of the general formula I have just stated.

Consider the unit circle. acts as the '' ordinate on the unit circle or in other words it is a parameter since it 'describes' all the '' values.

Now when does equal to ?

Surely, when the ordinate on the unit circle is equal to we have

So what is the ordinate when ? The first one you encounter is

What is the next one?

To generalise this: For all 'odd' coefficients of we have satisfied

Thus

[Hielly]

Hey for this Q
y=sin2(theta - pi)
Amp:1
Period:pi

the sin graph moves pi in the positive direction, so it starts from pi/2 on the x-axis? The answer says that its at the origin and touches the points pi/2 and pi.

Thanks

[Hielly]

How would i find the asymptotes for this y=3tan2(x-pi/4)-2 ?


Thanks

[Mao]

has asymptotes at

Hence, the asymptotes for your function would be at

[Hielly]

hey yeah thanks, but could you have used the general formula. Essentials states x=(2k+1)/2*pi/2 +b

not sure how to use this formula?


thanks

[TrueTears]

How would i find the asymptotes for this y=3tan2(x-pi/4)-2 ?


Thanks

General formula for the asymptotes for is

For

It is for all

So just sub in