# Edexcel C3 trigonometry

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

Anyone got any tips for the 'prove x = y' trigonometry questions? Just tried to do Q3 of June 2013 after studying it for most of the day and it had me completely stumped, I had to look at the mark scheme in the end.

Is there some kind of algorithm that people do to see what you're supposed to be factorising or cancelling or whatever?

Is there some kind of algorithm that people do to see what you're supposed to be factorising or cancelling or whatever?

0

reply

Report

#3

(Original post by

Anyone got any tips for the 'prove x = y' trigonometry questions? Just tried to do Q3 of June 2013 after studying it for most of the day and it had me completely stumped, I had to look at the mark scheme in the end.

Is there some kind of algorithm that people do to see what you're supposed to be factorising or cancelling or whatever?

**jf1994**)Anyone got any tips for the 'prove x = y' trigonometry questions? Just tried to do Q3 of June 2013 after studying it for most of the day and it had me completely stumped, I had to look at the mark scheme in the end.

Is there some kind of algorithm that people do to see what you're supposed to be factorising or cancelling or whatever?

Look at the available formulae

Think how can I get from this to this

Practice until you know how to get from this to this

0

reply

(Original post by

Let's see the question.

**Old_Simon**)Let's see the question.

2cos(x + 50) = sin(x + 40)

Show, without using a calculator, that

tan x = 1/3(tan 40)

0

reply

Also, could any tell me if I need to learn the factor formulae or will it be given in the booklet during the exam? Thanks

0

reply

Report

#7

(Original post by

Given that

2cos(x + 50) = sin(x + 40)

Show, without using a calculator, that

tan x = 1/3(tan 40)

**jf1994**)Given that

2cos(x + 50) = sin(x + 40)

Show, without using a calculator, that

tan x = 1/3(tan 40)

If you're still stuck, can you see how you could expand both sides using the formula's given in your formula book? From there you need to find a relationship between sin and cos in order to re-arrange for tan.

0

reply

Report

#8

**jf1994**)

Given that

2cos(x + 50) = sin(x + 40)

Show, without using a calculator, that

tan x = 1/3(tan 40)

2 [ cos(x)cos(50) - sin(x)sin(50) ] = sin(x)cos(40) + cos(x)sin(40)

Rearranging gives:

2cos(x)cos(50) - cos(x)sin(40) = sin(x)cos(40) + 2sin(x)sin(50)

A trig. identity you should know is that:

sin(x) = cos(90 - x)

and conversely:

cos(x) = sin(90 - x)

so cos(50) = sin(40) and cos(40) = sin(50)

So now we have:

2cos(x)sin(40) - cos(x)sin(40) = sin(x)cos(40) + 2sin(x)cos(40)

Therefore:

cos(x)sin(40) = 3sin(x)cos(40)

Divide by sin(x) * sin(40):

cos(x) / sin(x) = 3cos(40) / sin(40)

Then take the reciprocal of both sides! We know that tan(x) = sin(x) / cos(x) so...

sin(x) / cos(x) = (1/3) sin(40) / cos(40)

Which is equal to:

tan(x) = (1/3)tan(40)

!

3

reply

Report

#9

(Original post by

If you expand both sides using the formula from the formula book for sin(A+B) and cos(A+B) you get:

2 [ cos(x)cos(50) - sin(x)sin(50) ] = sin(x)cos(40) + cos(x)sin(40)

Rearranging gives:

2cos(x)cos(50) - cos(x)sin(40) = sin(x)cos(40) + 2sin(x)sin(50)

A trig. identity you should know is that:

sin(x) = cos(90 - x)

and conversely:

cos(x) = sin(90 - x)

so cos(50) = sin(40) and cos(40) = sin(50)

So now we have:

2cos(x)sin(40) - cos(x)sin(40) = sin(x)cos(40) + 2sin(x)cos(40)

Therefore:

cos(x)sin(40) = 3sin(x)cos(40)

Divide by sin(x) * sin(40):

cos(x) / sin(x) = 3cos(40) / sin(40)

Then take the reciprocal of both sides! We know that tan(x) = sin(x) / cos(x) so...

sin(x) / cos(x) = (1/3) sin(40) / cos(40)

Which is equal to:

tan(x) = (1/3)tan(40)

!

**Chazatthekeys**)If you expand both sides using the formula from the formula book for sin(A+B) and cos(A+B) you get:

2 [ cos(x)cos(50) - sin(x)sin(50) ] = sin(x)cos(40) + cos(x)sin(40)

Rearranging gives:

2cos(x)cos(50) - cos(x)sin(40) = sin(x)cos(40) + 2sin(x)sin(50)

A trig. identity you should know is that:

sin(x) = cos(90 - x)

and conversely:

cos(x) = sin(90 - x)

so cos(50) = sin(40) and cos(40) = sin(50)

So now we have:

2cos(x)sin(40) - cos(x)sin(40) = sin(x)cos(40) + 2sin(x)cos(40)

Therefore:

cos(x)sin(40) = 3sin(x)cos(40)

Divide by sin(x) * sin(40):

cos(x) / sin(x) = 3cos(40) / sin(40)

Then take the reciprocal of both sides! We know that tan(x) = sin(x) / cos(x) so...

sin(x) / cos(x) = (1/3) sin(40) / cos(40)

Which is equal to:

tan(x) = (1/3)tan(40)

!

2

reply

**Chazatthekeys**)

If you expand both sides using the formula from the formula book for sin(A+B) and cos(A+B) you get:

2 [ cos(x)cos(50) - sin(x)sin(50) ] = sin(x)cos(40) + cos(x)sin(40)

Rearranging gives:

2cos(x)cos(50) - cos(x)sin(40) = sin(x)cos(40) + 2sin(x)sin(50)

A trig. identity you should know is that:

sin(x) = cos(90 - x)

and conversely:

cos(x) = sin(90 - x)

so cos(50) = sin(40) and cos(40) = sin(50)

So now we have:

2cos(x)sin(40) - cos(x)sin(40) = sin(x)cos(40) + 2sin(x)cos(40)

Therefore:

cos(x)sin(40) = 3sin(x)cos(40)

Divide by sin(x) * sin(40):

cos(x) / sin(x) = 3cos(40) / sin(40)

Then take the reciprocal of both sides! We know that tan(x) = sin(x) / cos(x) so...

sin(x) / cos(x) = (1/3) sin(40) / cos(40)

Which is equal to:

tan(x) = (1/3)tan(40)

!

sin(x) = cos(90 - x)

and conversely:

cos(x) = sin(90 - x)

I didn't know that Are there any other trig identities I should know, other than the obvious ones? I'm self-taught and I've just been going off the examsolutions tutorials.

Thanks for your help

0

reply

Report

#11

(Original post by

A trig. identity you should know is that:

sin(x) = cos(90 - x)

and conversely:

cos(x) = sin(90 - x)

I didn't know that Are there any other trig identities I should know, other than the obvious ones? I'm self-taught and I've just been going off the examsolutions tutorials.

Thanks for your help

**jf1994**)A trig. identity you should know is that:

sin(x) = cos(90 - x)

and conversely:

cos(x) = sin(90 - x)

I didn't know that Are there any other trig identities I should know, other than the obvious ones? I'm self-taught and I've just been going off the examsolutions tutorials.

Thanks for your help

0

reply

Report

#12

**jf1994**)

A trig. identity you should know is that:

sin(x) = cos(90 - x)

and conversely:

cos(x) = sin(90 - x)

I didn't know that Are there any other trig identities I should know, other than the obvious ones? I'm self-taught and I've just been going off the examsolutions tutorials.

Thanks for your help

sin(-x) = -sin(x) also cosec(-x) = -cosec(x)

cos(-x) = cos(x) also sec(-x) = sec(x)

tan(-x) = -tan(x) also cot(-x) = -cot(x)

sin(x) = cos(90 - x)

cos(x) = sin(90 - x)

sin^2(x) = (1/2)(1 - cos(2x))

cos^2(x) = (1/2)(1 + cos(2x))

^^^ important for integration!

sin(2x) = 2sin(x)cos(x)

cos(2x) = cos^2(x) - sin^2(x) = 2cos^2(x) - 1 = 1 - 2sin^2(x)

tan(2x) = 2tan(x) / ( 1 - tan^2(x) )

How to right:

Asin(x) + Bcos(x) in the form Rsin(x + a)

sin^2(x) + cos^2(x) = 1

this implies:

(divide by sin^2(x)): 1 + cot^2(x) = cosec^2(x)

(divide by cos^2(x)): tan^2(x) + 1 = sec^2(x)

^^^ important!

Also the ones in the formula book for sin(A+B) and cos(A+B) and tan(A+B) are also important, and definitely the ones below like:

sin(A) + sin(B)

sin(A) - sin(B)

cos(A) + cos(B) and

cos(A) - cos(B) are all very important, both for integrating and solving equations:

Having it in the form sin(A) + sin(B) is best for integration and in the form:

2sin[ (A+B)/2 ]cos[ (A-B)/2 ] for solving equations!

2

reply

Report

#13

(Original post by

Try not to post full solutions in the future, it's a lot better to help people to work it out themselves

**lmorgan95**)Try not to post full solutions in the future, it's a lot better to help people to work it out themselves

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top