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Very hard C3 Trig Question! Help..

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1. Very hard C3 Trig Question! Help..
Solve, for 0◦< β< 90◦

the equation 3 sin 6β cosec 2β = 4.

..

What i did..

First i know cosec 2β is the same thing as 1/sin2β

I made sin6β into = 2sin3βcos3β

So now i have 3(2sin3βcos3β)x1/sin2β = 4
-> 6sin3βcos3βx1/sin2β = 4
--> sin3βcos3β = 2sin2β/3

..

I tried taking it one side and i just can't figure out what to do next..

Thanks.

Jan 06 Question 9iii) http://blogs.thegrangeschool.net/mat...er-Booklet.pdf
Last edited by Abed1993; 13-06-2012 at 15:33.
2. Re: Very hard C3 Trig Question! Help..
(Original post by notnek)
Instead of sin(6b) = sin(3b+3b), try using sin(6b)=sin(4b+2b).
I'll try it out..
3. Re: Very hard C3 Trig Question! Help..
how many marks and where is this question from?
4. Re: Very hard C3 Trig Question! Help..
use sin(4x+ 2x)

then do it again as sin(2x + 2x) on the 4x terms.
5. Re: Very hard C3 Trig Question! Help..
(Original post by Alpha5)
how many marks and where is this question from?
The link is on the description

and it's 6 marks
6. Re: Very hard C3 Trig Question! Help..
(Original post by Ilyas)
use sin(4x+ 2x)

then do it again as sin(2x + 2x) on the 4x terms.
I used sin(4x+ 2x)

and got sin4xcos2x+cos4xsin2x
then multiplied by 3
and got 3sin4xcos2x+3cos4xsin2x

Now i should do it again but for sin(2x+2x)?

EDIT:

I now have 3sin2xcos^2(2x) + 3cos^2(2x)sin2x + 3cos4xsin2x :/

You sure this is right? Very tedious..
Last edited by Abed1993; 13-06-2012 at 15:38.
7. Re: Very hard C3 Trig Question! Help..
(Original post by Abed1993)
I used sin(4x+ 2x)

and got sin4xcos2x+cos4xsin2x
then multiplied by 3
and got 3sin4xcos2x+3cos4xsin2x

Now i should do it again but for sin(2x+2x)?

EDIT:

I now have 3sin2xcos^2(2x) + 3cos^2(2x)sin2x + 3cos4xsin2x :/

You sure this is right? Very tedious..
change the cos4x into cos(2x + 2x), basically eliminate all 4x terms and see what happens
8. Re: Very hard C3 Trig Question! Help..
The first part of the question proves an identity for so the simplest method is to use this to form an identity for .

Do you have any ideas how to do this?
Last edited by notnek; 13-06-2012 at 15:43.
9. Re: Very hard C3 Trig Question! Help..
(Original post by notnek)
The first part of the question proves an identity for so the simplest method is to use this to form an identity for .

Do you have any ideas how to do this?
Oh yes, i didn't think of using previous answers..

sin3x to sin6x..
can't i just make x=3x and sub it in..
10. Re: Very hard C3 Trig Question! Help..
(Original post by Abed1993)
Oh yes, i didn't think of using previous answers..

sin3x to sin6x..
can't i just make x=3x and sub it in..
x=3x will give you an identity for sin(9x). x=2x should work better.
11. Re: Very hard C3 Trig Question! Help..
(Original post by notnek)
x=3x will give you an identity for sin(9x). x=2x should work better.
12. Re: Very hard C3 Trig Question! Help..
lol, if you went the first route it would have taken you a very long time, what ive noticed about these questions is that when they first tell you to identify an identity, they always have a following questions where you will need to sub something in. hope that helps but im doing edexcel might be different.
13. Re: Very hard C3 Trig Question! Help..
(Original post by Abed1993)
Solve, for 0◦< β< 90◦

the equation 3 sin 6β cosec 2β = 4.
Arranging the equation and :

sinA=/=0 because A=/=0 so

And solve for A then

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Last updated: June 14, 2012
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