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Radian/trig A level mathematics help

Basically I'm stuck, my teacher hasn't taught this far yet but I'm eager to know how and the method just won't come to my head. Any help with 9ai would be great, I can attempt to apply the method to others.

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Reply 1

Notnek

Original post by CameronWS

Do you know about the "CAST" diagram or do you generally only use the trig graphs?

Reply 2

CameronWS

Original post by Sir Cumference

Do you know about the "CAST" diagram or do you generally only use the trig graphs?

The who what now? Just trig graphs/triangles sort of thing.

Reply 3

GrayestOwl0900

Pi radians = 180 degrees.

Go from here and manipulate the sides to get what you want.

Use the properties of sine and cosine angles. Like what Sir Cumference said, you can use the graphs of y = sinx and
y = cosx to identify these properties (or just know them off by heart)

(edited 4 years ago)

Reply 4

Notnek

Original post by CameronWS

The who what now? Just trig graphs/triangles sort of thing.

Ok ignore CAST - I was just checking how you're being taught in school. Have a look at this:

https://www.desmos.com/calculator/vxptxjg5yk

How are

\cos \frac{3 \pi}{4}

and

\cos \frac{\pi}{4}

related? And do you know the value of

\cos \frac{\pi}{4}

?

For all of these questions, you need to use symmetry of the trig graphs to express the required value in terms of sin/cos/tan of an acute angle like

\frac{\pi}{4}

for example.

(edited 4 years ago)

Reply 5

0le

A 360° revolution of a circle has

2\pi

radians. You can use this idea to work out the answers, provided you know basic rules such as

\cos(0) = 1

and the diagram below:

https://en.wikipedia.org/wiki/Unit_circle#/media/File:Circle-trig6.svg

Reply 6

CameronWS

Original post by Sir Cumference

Ok igore CAST - I was just checking. Have a look at this:

https://www.desmos.com/calculator/vxptxjg5yk

How are

\cos \frac{3 \pi}{4}

and

\cos \frac{\pi}{4}

related? And do you know the value of

\cos \frac{\pi}{4}

?

For all of these questions, you need to use symmetry of the trig graphs to express the required value in terms of sin/cos/tan of an acute angle.

I can't use a calculator, and I can't use desmos in an exam. I don't know how they're related either. :frown:

Reply 7

danishh0007

Multiply each radian by 180/pi to get it in terms of degrees and use whatever method to find exact trig value

Reply 8

Notnek

Original post by CameronWS

I can't use a calculator, and I can't use desmos in an exam. I don't know how they're related either. :frown:

I know you can't use Desmos - I was just illustrating an example but you will need to draw the graphs yourself once you understand the general method. Can you not see that by the symmetry of the graph,

\cos \frac{\pi}{4}

and

\cos \frac{3\pi}{4}

are equal but one is positive and one is negative?

You should have learnt what

\cos \frac{\pi}{4}

is at GCSE. Do you remember it?

Reply 9

CameronWS

Original post by Sir Cumference

\cos \frac{\pi}{4}

and

\cos \frac{3\pi}{4}

are equal but one is positive and one is negative?

You should have learnt what

\cos \frac{\pi}{4}

is at GCSE. Do you remember it?

Oh I thought the symmetry of them was common knowledge. (So 3pi/4 = pi - pi/4?)
Yes cos(pi/4) = root2/2.

Apparently there is a triangular method?

(edited 4 years ago)

Reply 10

Notnek

Original post by CameronWS

Oh I thought the symmetry of them was common knowledge. (So 3pi/4 = pi - pi/4?)
Yes cos(pi/4) = root2/2.

Apparently there is a triangular method?

Correct so that means that

\cos \frac{3\pi}{4}

must be

-\frac{\sqrt{2}}{2}

. Try the others now by drawing a graph and trying to express the values in terms of acute angles like I did with

\frac{\pi}{4}

.

Do you mean a triangle method for finding e.g.

\cos \frac{\pi}{4}

- the ones learnt at GCSE?

Reply 11

CameronWS

Original post by Sir Cumference

Correct so that means that

\cos \frac{3\pi}{4}

must be

-\frac{\sqrt{2}}{2}

. Try the others now by drawing a graph and trying to express the values in terms of acute angles like I did with

\frac{\pi}{4}

.

Do you mean a triangle method for finding e.g.

\cos \frac{\pi}{4}

- the ones learnt at GCSE?

I'm really sorry I simply just don't understand how what I did means it is equal to - root2/2.

Reply 12

Notnek

Original post by CameronWS

I'm really sorry I simply just don't understand how what I did means it is equal to - root2/2.

Oh I thought when you said about it was "common knowledge" that you understood the symmetry of the cos graph. Look at this diagram:

https://www.desmos.com/calculator/rr43ijdpnx

Can you see that

\cos \frac{3\pi}{4} = -\cos \frac{\pi}{4}

i.e. they are the same value but just different signs?

Reply 13

CameronWS

Original post by Sir Cumference

Oh I thought when you said about it was "common knowledge" that you understood the symmetry of the cos graph. Look at this diagram:

https://www.desmos.com/calculator/rr43ijdpnx

Can you see that

\cos \frac{3\pi}{4} = -\cos \frac{\pi}{4}

i.e. they are the same value but just different signs?

I think I should add the fact that I have dyscalculia and visualising/reading graphs is next to impossible for me, I'm really trying.
I need a more mathematical understanding of how to solve it than thinking of graphs, if you get me?

Reply 14

CameronWS

Original post by Sir Cumference

Oh I thought when you said about it was "common knowledge" that you understood the symmetry of the cos graph. Look at this diagram:

https://www.desmos.com/calculator/rr43ijdpnx

Can you see that

\cos \frac{3\pi}{4} = -\cos \frac{\pi}{4}

i.e. they are the same value but just different signs?

Okay so tell me if I'm wrong but cos(3pi/4) = pi - pi/4.
We know 1piRad/4 = 45. Such that cos(45) = root2/2. (As we know as common knowledge) Since there's a shift of 1piRadian (pi - pi/4) it's root2/2 but negative?

Reply 15

Notnek

Original post by CameronWS

I think I should add the fact that I have dyscalculia and visualising graphs is next to impossible for me, I'm really trying.
I need a more mathematical understanding of how to solve it than thinking of graphs, if you get me?

If you could explain what you're seeing in the diagram above I can guide you. I recommend trying this:

Put your finger on

\pi/4

on the x-axis and move up to the cos graph. Then turn left and move until you reach the y-axis. This is the value of

\cos \pi/4

. Now do the same thing with

3\pi/4

on the x-axis.

You really need to see that by symmetry, the values that you've reached on the y-axis are the same distance from the origin. In other words one is the negative of the other.

Trying to do these questions without using the graph is going to be tedious and to be honest it may be a waste of your time since you'll have a calculator in the real exam. Without the graphs you could just learn all the rules e.g.

\cos (\pi - x) = -\cos(x)

\cos 3\pi/4 = -\cos \pi/4

.

I think you need to try harder to use the graph and not give up so quickly. I bet you're just not understanding what I'm trying to show you and you can see the symmetry if I could explain one-to-one.

Reply 16

Notnek

Original post by CameronWS

But why is it negative? Would this also work with sin/tan?

If you do know all the rules for sin/cos/tan off by hear then that's fine - please let me know.

Reply 17

CameronWS

Original post by Sir Cumference

But why is it negative? Would this also work with sin/tan?

If you do know all the rules for sin/cos/tan off by hear then that's fine - please let me know.

It's negative because with a cos graph by shifting by 1 pi radian I'm pretty certain but not 100% but you get the same value just negative/positive?
I don't know all the rules no.

Thanks for all your help so far.

Reply 18

Notnek

Original post by CameronWS

What you're saying is a bit confusing because you're not shifting by pi radians, instead you're subtracting the value from pi. The rule in degrees (to make it easier) is

\cos(180 - x) = -\cos(x)

E.g.
cos(160) = -cos(20)
cos(130) = -cos(50)

This rule can be seen using the graph and you can get everything else from the sin/cos/tan graphs - you seem reluctant to though. I recommend asking your teacher about these questions because it seems like you really need one-to-one guidance.

Reply 19

CameronWS

Original post by Sir Cumference

What you're saying is a bit confusing because you're not shifting by pi radians, instead you're subtracting the value from pi. The rule in degrees (to make it easier) is