Addition formulae - P2

Please help me with following:

cos^2 (pi/2 + x) + sin^2 (pi/2 - x)

This is a bit different and I don't know how to get round it.

Help...

Thank you

Jorge

Reply 1

Gauss

Jorge

Please help me with following:

cos^2 (pi/2 + x) + sin^2 (pi/2 - x)

This is a bit different and I don't know how to get round it.

Help...

Thank you

Jorge

Let X = pi/2 + x
Let Y = pi/2 - x

cos^2(X) = [1 + cos2X]/2
sin^2(Y) = [1 - cos2Y]/2

Can you pick it up from there?

Euclid.

Reply 2

RichE

Jorge

Please help me with following:

cos^2 (pi/2 + x) + sin^2 (pi/2 - x)

This is a bit different and I don't know how to get round it.

Help...

Thank you

Jorge

If you use

sin(pi/2-x) = cos x

and

cos(pi/2 - (-x)) = cos(-x) = cos x

then it simplifies somewhat more quickly.

Reply 3

Jorge

Thanks

I forgot to say that what the exercice wants is

to simplify

The answer is 1, and I'm afraid I can't get there,

Let X = pi/2 + x
Let Y = pi/2 - x

cos^2(X) = [1 + cos2X]/2
sin^2(Y) = [1 - cos2Y]/2

Can you pick it up from there?

Euclid.

cos^2 x + sin^2 y

1 - sin^2 x + sin^2 y

sin^2 (xy) =1

But I don't think this is what they want...

Help...

Jorge

Reply 4

Gauss

Jorge

Thanks

I forgot to say that what the exercice wants is

to simplify

The answer is 1, and I'm afraid I can't get there,

cos^2 x + sin^2 y

1 - sin^2 x + sin^2 y

sin^2 (xy) =1

But I don't think this is what they want...

Help...

Jorge

Let X = pi/2 + x
Let Y = pi/2 - x

cos^2(X) = [1 + cos2X]/2
sin^2(Y) = [1 - cos2Y]/2

=> cos^2(X) + sin^2(X) = [1 + cos2X]/2 + [1 - cos2Y]/2
=> [1 + cos2(pi/2 + x)]/2 + [1 - cos2(pi/2 - x)]/2
=> [1 + cos(pi + 2x)]/2 + [1 - cos(pi - 2x)]/2
=> [1 + cos(pi)cos2x - sin(pi)sin2x]/2 + [1 - cos(pi)cos2x + sin(pi)sin2x]/2

you know that cos(pi) = -1, and sin(pi) = 0

=> [1 - cos2x]/2 + [1 + cos2x]/2
=> sin^2(x) + cos^2(x)
=> 1

Euclid.

Reply 5

Jorge

Euclid

Are you using the double angle formulae?

Because this exercice is prior to that, and I'm still a bit lost.

Jorge

Reply 6

Gauss

Jorge

Euclid

Are you using the double angle formulae?

Because this exercice is prior to that, and I'm still a bit lost.

Jorge

In that case it is probably best to use the fact that cos(pi/2 + x) = -sinx and sin(pi/2 - x) = cosx.

Euclid.

Reply 7

Jorge

Hello Euclid

Thanks for your help

I'm still wondering as to the route the book wants you to follow.

At this stage - the addition formulae - I think that either I use it, or take advantage of cos^2A + sin^2A = 1

Now, as the result is simply: 1, I have a feeling that is what they're after.

Except that I am not too sure about the contents of the parentheses.

I suppose that as you are multiplying neg * pos you probably end up with nothing.

Could youy help me along these lines?

(This is a problem on page 57 of Collin's Pure 2, for those who might have the book)

Thanks

Jorge

Reply 8

Gauss

Jorge

OK., what identities are you allowed to use at this stage for answering that question? Then we'll worth with them.

Euclid.

Reply 9

Jorge

I guess the addition fomulae and cos^2A + sin^2A = 1

From other questions in the same group it looks like they want me to know that things like cos pi/3 = 0.5, etc.. which makes life difficult for me.

Jorge

Reply 10

Gauss

Jorge

I'm very sorry but remind me what the addition fomulae formuale are. I never remember the names :redface:

Euclid.

Reply 11

Jorge

Euclid

I'm very sorry but remind me what the addition fomulae formuale are. I never remember the names :redface:

Euclid.

Same problem here!

Addtion formulae:

sin(A+B) = sinAcosB + sinBcosA

rings a bell?

Jorge

Reply 12

Gauss

Jorge

Same problem here!

Addtion formulae:

sin(A+B) = sinAcosB + sinBcosA

rings a bell?

Jorge

You have: cos^2 (pi/2 + x) + sin^2 (pi/2 - x)

The using the addition formula:

cos(pi/2 + x) = cos[pi/2]cosx - sin[pi/2]sinx
sin(pi/2 - x) = sin[pi/2]cos(-x) + cos[pi/2]sin(-x)

using the fact that cos90 = 0, and sin90 = 1, the equation reduces to:

cos(pi/2 + x) = cos[pi/2]cosx - sin[pi/2]sinx = -sinx
sin(pi/2 - x) = sin[pi/2]cos(-x) + cos[pi/2]sin(-x) = cos(-x) = cosx

You are then left with:

(-sinx)^2 + (cosx)^2

which you know is equal to one.

Euclid.

Reply 13

Jorge

Euclid

Ahh right, they used to call it the double angle formulae when I was doing it.

You have: cos^2 (pi/2 + x) + sin^2 (pi/2 - x)

The using the addition formula:

cos(pi/2 + x) = cos[pi/2]cosx - sin[pi/2]sinx
sin(pi/2 - x) = sin[pi/2]cos(-x) + cos[pi/2]sin(-x)

using the fact that cos90 = 0, and sin90 = 1, the equation reduces to:

cos(pi/2 + x) = cos[pi/2]cosx - sin[pi/2]sinx = -sinx
sin(pi/2 - x) = sin[pi/2]cos(-x) + cos[pi/2]sin(-x) = cos(-x) = cosx

You are then left with:

(-sinx)^2 + (cosx)^2

which you know is equal to one.

Euclid.

Thanks Euclid

What's confusing me here is the use of squares.

The additition formulae are for cos(A+B) and here is cos^(A+B)

Am I to understand that you proceed in the same fashion as if it were not squared and simply square the result at the end?

Thanks

Jorge

Reply 14

Gauss

Jorge

Yes because

cos^2(A+B) = cos(A+B)*cos(A+B)

Do one of them and then square it.

Euclid.

Reply 15

Jorge

Euclid

Yes because

cos^2(A+B) = cos(A+B)*cos(A+B)

Do one of them and then square it.

Euclid.

Ahhhhh.....

Thanks

Jorge

Reply 16

Newton

((cos((Pi/2)+x))^2)+((sin((Pi/2)-x))^2)

cos((Pi/2)+x)=cos(Pi/2)cosx-sin(Pi/2)sinx=-sinx
sin((Pi/2)-x)=sin(Pi/2)cosx-sinxcos(Pi/2)=cosx

((cos((Pi/2)+x))^2)+((sin((Pi/2)-x))^2)=((-sinx)^2)+((cosx)^2)=((sinx)^2)+((cosx)^2)

=1 Q. E. D.

Newton.