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# Addition formulae - P2 watch

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
2. (Original post by Jorge)

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.
3. (Original post by Jorge)

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.
4. 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
5. (Original post by 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.
6. Euclid

Are you using the double angle formulae?

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

Jorge
7. (Original post by 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.
8. Hello Euclid

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
9. (Original post by Jorge)
Hello Euclid

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

Euclid.
10. 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
11. (Original post by 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
I'm very sorry but remind me what the addition fomulae formuale are. I never remember the names

Euclid.
12. (Original post by Euclid)
I'm very sorry but remind me what the addition fomulae formuale are. I never remember the names

Euclid.

Same problem here!

sin(A+B) = sinAcosB + sinBcosA

rings a bell?

Jorge
13. (Original post by Jorge)
Same problem here!

sin(A+B) = sinAcosB + sinBcosA

rings a bell?

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

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.
14. (Original post by 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)

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
15. (Original post by Jorge)
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
Yes because

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

Do one of them and then square it.

Euclid.
16. (Original post by 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
17. ((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.

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