[Abed1993]

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

[Abed1993]

(Original post by notnek)
Instead of sin(6b) = sin(3b+3b), try using sin(6b)=sin(4b+2b).

I'll try it out..

[Alpha5]

how many marks and where is this question from?

[Venomilys]

use sin(4x+ 2x)

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

[Abed1993]

(Original post by Alpha5)
how many marks and where is this question from?

The link is on the description

and it's 6 marks

[Abed1993]

(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..

[Venomilys]

(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

[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?

[Abed1993]

(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..

[notnek]

(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.

[Abed1993]

(Original post by notnek)
x=3x will give you an identity for sin(9x). x=2x should work better.

Yep sorry my bad.

[Abdullah CEO]

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.

[ztibor]

(Original post by Abed1993)
Solve, for 0◦< β< 90◦

the equation 3 sin 6β cosec 2β = 4.

Arranging the equation and :

with addition rule



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

And solve for A then