8((sin2a)^2)-4 as a single function of sin(a) or cos(a)

 

sry Chapter final i had no idea about and this was the only one that slipped me.

by the way anyone know how to put say y^2 so that it looks like y squared?

y²

That's using [ sup] tags (without the space).

Or

y²

alt 178

Using the formula that Sin2(a) = 2Sin(a)Cos(a)

 

Bring the 2 outside the bracket and multiply by the 8:

16[(Sin(a)Cos(a))^2]-4

 

Then you multiply the Sin(a)Cos(a) by itself and this changes the formula to:

16[sin^2(a) + 2Sin(a)Cos(a) + Cos^2(a)]-4

 

Since Sin^2(a) = 1- Cos^2(a) The formula changes to:

16[1- Cos^2(a) + 2Sin(a)Cos(a) + Cos^2(a)]-4

 

The two Cos^2(a) cancels leaving:

16[1 - 2Sin(a)Cos(a)]-4

 

Multply out and you get:

16 - 32Sin(a)Cos(a) -4

 

Leaving you with an answer of:

12 - 32Sin(a)Cos(a)

 

 

I hope this makes sense to you because as you may have figured out, i don't know how to put in ^2, which would make it easier to read.

 

Anyway if you don't understand it, just say the word and i will happily try and re-do it in a better format.

hmm i get that and thank you, but how do you put it as just a sin(a) function

 

THank you

oops, thought it was sina and cosa! i'll get back to you

2Sin(a)Cos(a) = Sin^2(a)

 

So the formula at the end becomes:

 

16 - 16(Sin^2(a))

 

How's that?

 

If you want to put it in terms of cos then:

16(1-Sin^2(a)) = 16 Cos^2(a)

2Sin(a)Cos(a) = Sin^2(a)

 

So the formula at the end becomes:

 

16 - 16(Sin^2(a))

 

How's that?

 

If you want to put it in terms of cos then:

16(1-Sin^2(a)) = 16 Cos^2(a)

 

hey i thought it was 2sin(a)cos(a)= Sin2(a)

Indeed. [math]2\sin(a)\cos(a) = \sin(2a)[/math]. Substitute this directly into your formula and use the fact that [math]\cos^2(x) = 1 - \sin^2(x)[/math].

hey i thought it was 2sin(a)cos(a)= Sin2(a)

 

 

You are correct. my mistake. must look into it a bit further so!

Try and come up with some suggestions yourself if you can.

Indeed. [math]2\sin(a)\cos(a) = \sin(2a)[/math]. Substitute this directly into your formula and use the fact that [math]\cos^2(x) = 1 - \sin^2(x)[/math'].

 

 

That would work except that he is looking for it in terms of Sin(a) and not Sin (2a).

I've really gotta learn how to put in those fancy formulas!

That would work except that he is looking for it in terms of Sin(a) and not Sin (2a).

 

The method I gave will give it in terms of sin(a).

But if you sub in your sin(2a) into the formula I worked out then you would get:

 

16-16Sin(2a).

 

Anyway it's too late for this (3a.m.!!), I'm going to bed!

8((sin2a)^2)-4

8(sin2(a))(sin2(a))-4

8(2sin(a)cos(a))(2sin(a)cos(a))-4

32(sin^2(a))(cos^2(a))-4

32sin^2(a)(1-sin^2(a))-4

32sin^2(a)-32sin^4(a)-4

4(8sin^2(a)-8sin^4(a)-1)

 

now what pls

 

ps. just a sin(a) or cos(a) funtion not to any power

You can try factoring it, but you're probably not going to remove the sin². I'm afraid I don't have time to look at this problem further atm; I shall look at it in a few hours time.

thank you

BAH! it's 4cos4a

 

(take out a negative four, then use pythagorean therom!)

 

thanks for the help by the way