Urgent trigonometry help please :)

Hi guys,
My maths tutor set me lots of questions to do as part of revision for this week. I'm stuck on one of the questions but she is on holiday so I can't ask her for help.
So please could someone kindly show me how to do the question in the attachment as I really don't understand it.

Thanks in advance to anyone to helps :smile:

Original post by Aty100

Hi guys,
My maths tutor set me lots of questions to do as part of revision for this week. I'm stuck on one of the questions but she is on holiday so I can't ask her for help.
So please could someone kindly show me how to do the question in the attachment as I really don't understand it.

Thanks in advance to anyone to helps :smile:

That tan looks out of place so you want to get rid of that first:

\displaystyle \frac{t\times s}{1-c} = \frac{\frac{s}{c} \times s}{1-c} = \frac{s^2}{c(1-c)}

What could you do next?

(edited 8 years ago)

Reply 2

Aty100

OP

15

Original post by notnek

That tan looks out of place so you want to get rid of that first:

\displaystyle \frac{t\times s}{1-c} = \frac{\frac{s}{c} \times s}{1-c} = \frac{s^2}{c(1-c)}

What could you do next?

Thank you. That really helped.
Also I don't completely understand where the 1+ comes from :/

(edited 8 years ago)

Reply 3

OrionMusicNet

13

Original post by Aty100

Thank you. That really helped. I'm not sure whether I did the last bit correctly. If you don't mind could you please check.
Also I don't completely under where the 1+ comes from :/

sin^2(theta) is the same as 1-cos^2(theta). Do you remember the difference of two squares rule?

Reply 4

Aty100

OP

15

Original post by OrionMusicNet

sin^2(theta) is the same as 1-cos^2(theta). Do you remember the difference of two squares rule?

I think I got that bit. Not sure if I did it correctly.

Original post by Aty100

I think I got that bit. Not sure if I did it correctly.

It should be

\displaystyle \frac{1-\cos^2 \theta}{\cos\theta \left(1-\cos \theta \right)}

You have a negative sign on the denominator. Typo?

Next step using difference of two squares:

\displaystyle \frac{(1+\cos \theta)(1-\cos \theta)}{\cos\theta \left(1-\cos \theta \right)}

Can you see what to do next?

Reply 6

OrionMusicNet

13

Original post by Aty100

I think I got that bit. Not sure if I did it correctly.

1-c is the same 1 - cos(theta). If you multiply top and bottom of the fraction by cos(theta) you get sin(theta) x sin(theta) on the top, and on the bottom you get cos(theta) - cos^2(theta). You can then factorise the bottom to get cos(theta)(1 - cos(theta)). You can see that if you expand that you would get back to cos(theta) - cos^2(theta). If you change the sin^2(theta) to 1 - cos^2(theta) you have a line in the form of the difference of two squares (a^2 - b^2 is the same as (a-b)(a+b)), therefore 1 - cos^2(theta) is the same as (1 - cos(theta))(1+cos(theta)) as 1^2 is simply 1. Can you see that the (1-cos(theta)) on the top will cancel with the (1-cos(theta) on the bottom of the fraction? You then simply need to split the fraction apart and you should see that you will get the answer.