jf1994
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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? :mad:
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Old_Simon
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Let's see the question.
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forsparta
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(Original post by 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? :mad:
Look at the end result you are trying to reach

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
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jf1994
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(Original post by Old_Simon)
Let's see the question.
Given that

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


Show, without using a calculator, that


tan x = 1/3(tan 40)
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jf1994
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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
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Old_Simon
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All required formulas are stated in the spec for each module.
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lmorgan95
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(Original post by jf1994)
Given that

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


Show, without using a calculator, that


tan x = 1/3(tan 40)
Don't worry, i don't know anyone that got that right in the exam lol.
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.
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Chazatthekeys
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(Original post by jf1994)
Given that

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


Show, without using a calculator, that


tan x = 1/3(tan 40)
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)

!
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lmorgan95
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(Original post by 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)

!
Try not to post full solutions in the future, it's a lot better to help people to work it out themselves
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jf1994
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(Original post by 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)

!
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
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Old_Simon
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(Original post by 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
Exam Solutions is good for answering questions but you really need text books or online web sites etc to go with it. Trig Identities are really the core of trig and the one used here where cos x = sin 90 - x , is just so basic. There are a whole mess of identities which are essential to know.
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Chazatthekeys
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(Original post by 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
Umm... Here are the ones that aren't in the formula book that are quite important, some are simple:

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!
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Chazatthekeys
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(Original post by lmorgan95)
Try not to post full solutions in the future, it's a lot better to help people to work it out themselves
sorry aha!
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