spontaneous test help!

[Asimov Pupil]

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.

[Asimov Pupil]

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

[Cap'n Refsmmat]

y²

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

Or

y²

alt 178

[dan19_83]

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.

[Asimov Pupil]

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

 

THank you

[dan19_83]

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

[dan19_83]

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)

[Asimov Pupil]

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)

[Dave]

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].

[dan19_83]

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.

[dan19_83]

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!

[Dave]

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).

[dan19_83]

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!

[Asimov Pupil]

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

[Dave]

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.

[Asimov Pupil]

thank you

[Asimov Pupil]

BAH! it's 4cos4a

 

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

 

thanks for the help by the way