Prove cos2(x)+sin2(x) = 1

Haha. I love stuff like that. Kind of reminds me of how to derive the series expansion for e^x (with ever so slight abuse of notation):

\int_0^x e^x dx= e^x - 1

\int_0^x e^x dx - e^x = -1

(\int_0^xdx - 1)e^x = -1

e^x = \frac{1}{1 - \int_0^xdx} = 1 + \int_0^xdx + \int_0^x\int_0^xdxdx + \int_0^x\int_0^x\int_0^xdxdxdx + ...

e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + ...

Nothing wrong with that at all...

Reply 81

IrrationalNumber

12

We integrate

(c^2+s^2 - 1)^2

over [0,pi/2]. Define the linear functional

I(f) = \int_0^{\frac{\pi}{2}} f(x)dx

Notice that

I(e^{4ix})=0

. Recalling that

e^{ix}=\cos(x) + i\sin(x)

, we see that

I(c^4 + s^4)=I(6c^2s^2)

. (you can do this without euler's identity using integration by parts)
Hence

I[(c^2+s^2-1)^2]= I(8c^2s^2 -2s^2 - 2c^2 +1)

. Now we use the double angle formulae to see that this is equal to

I(2(\sin^2(2x) -s^2) - \cos(2x))

We can see that

Unparseable latex formula:

I(2(\sin^2(2x)) = \int_0^{\pi}} \sin^2(x) dx

. By the symmetry of sine about pi/2, we see that

I(2(\sin^2(2x))=2I(\sin^2(x))

Finally, it is easy to see that the integral over [0,pi] of cos(x) is 0. This leaves us with

I((c^2+s^2 - 1)^2) =0

. Using continuity of sine and cosine we see that c^2+s^2=1 on [0,pi/2]. But then they are equal everywhere because sin^2(pi-x) = sin^2(x) and cos^2(pi-x) = cos^2(x) (and so they are equal on [pi/2,pi]) and s^2+c^2 is a pi periodic function.

(edited 12 years ago)

Reply 82

joe17292

Original post by tonyhawken

It may constitute a proof, but is a very roundabout way of doing things. Also note if you believe some of the other posters on this forum, it would not constitute a real proof as you have not gone right back to basics - you may have made too many assumptions.

Maybe you haven't realised what the point of this thread is?! It's to find as many ways as possible to prove the equation. So roundabout proofs are pretty much exactly what it's asking for. And yes I haven't included the proof that the derivative of sin(x) is cos(x) and the derivative of cos(x) and -sin(x), because I expect the reader to already know that that is true... :bhangra:

Reply 83

tonyhawken

10

Original post by Rahul.S

i had another way of proving this....but i must say i really like this way...afterall im learning de moivre's theorem :colone:

And this is a neat and efficient proof.

Reply 84

falcon pluse

Im going to be starting A level maths/fm , is it ok that i dont know this?

Reply 85

joe17292

Original post by falcon pluse

Im going to be starting A level maths/fm , is it ok that i dont know this?

When you say 'this', do you mean the fact that sin^2(x)+cos^2(x)=1 ?

Reply 86

falcon pluse

Original post by joe17292

When you say 'this', do you mean the fact that sin^2(x)+cos^2(x)=1 ?

noo , being able to prove it . I know that identity

Reply 87

joe17292

Original post by falcon pluse

noo , being able to prove it . I know that identity

Ok. I wouldn't say you need to have learnt it by now, but everyone should know how to prove it by the time they start using it (which you will be doing in AS maths). Also out of interest, do you know any proofs of Pythagoras' theorem?

Reply 88

falcon pluse

Original post by joe17292

Ok. I wouldn't say you need to have learnt it by now, but everyone should know how to prove it by the time they start using it (which you will be doing in AS maths). Also out of interest, do you know any proofs of Pythagoras' theorem?

Ive done AS maths and furthermaths lol , we were taught but i didnt pay much attention to the proofs so no i do not :biggrin:

( im not bad at maths so if you have an area which is an easy way to see how to do it ill have a look as im applying for maths with economics at uni )

Reply 89

Rahul.S

14

Original post by tonyhawken

And this is a neat and efficient proof.

yh defo

Reply 90

Aristotle's' Disciple

19

Sick.

Reply 91

Oh my Ms. Coffey

Original post by L'art pour l'art

\begin{aligned}1 & = \cos(0) \\& = \cos(x-x) \\& = \cos{x}\cos{x}+\sin{x}\sin{x} \\& = \cos^2{x}+\sin^2{x}.\end{aligned}

Love it, I can actually understand that one :teehee:

Reply 92

han87xx

My favourite simple proof, probably been posted

cos(x) = adjacent/hypotenuse

sin(x) = opposite/hypotenuse

cos2(x) +sin2(x) = ((adjacent)^2 + (opposite)^2)/(hypotenuse)^2

By Pythagoras, (adjacent)^2 + (opposite)^2 = (hypotenuse)^2

so ((adjacent)^2 + (opposite)^2)/(hypotenuse)^2 = (hypotenuse)^2/(hypotenuse)^2 = 1

Reply 93

Xero Xenith

10

Original post by han87xx

My favourite simple proof, probably been posted

cos(x) = adjacent/hypotenuse

sin(x) = opposite/hypotenuse

cos2(x) +sin2(x) = ((adjacent)^2 + (opposite)^2)/(hypotenuse)^2

By Pythagoras, (adjacent)^2 + (opposite)^2 = (hypotenuse)^2

so ((adjacent)^2 + (opposite)^2)/(hypotenuse)^2 = (hypotenuse)^2/(hypotenuse)^2 = 1

That's my favourite way to explain it to people, it's good :smile:

but I think it's pretty much identical to post #2?

Reply 94

Rahul.S

14

Original post by han87xx

My favourite simple proof, probably been posted

cos(x) = adjacent/hypotenuse

sin(x) = opposite/hypotenuse

cos2(x) +sin2(x) = ((adjacent)^2 + (opposite)^2)/(hypotenuse)^2

By Pythagoras, (adjacent)^2 + (opposite)^2 = (hypotenuse)^2

so ((adjacent)^2 + (opposite)^2)/(hypotenuse)^2 = (hypotenuse)^2/(hypotenuse)^2 = 1

this is the proof i used before.....but some of the proofs here are quite unique and less lenghty

Reply 95

Zuzuzu

12

Hmm...

\displaystyle \sin ^2 x = (2 \sin \frac{x}{2} \cos \frac{x}{2})^2

[INDENT]

\displaystyle = (2 \sin ^2 \frac{x}{2})(2 \cos ^2 \frac{x}{2})

\displaystyle = (1 - \cos x)(1 + \cos x)

\displaystyle = 1 - \cos^2 x

[/INDENT]

Reply 96

RichE

15

Original post by Zuzuzu

Hmm...

\displaystyle \sin ^2 x = (2 \sin \frac{x}{2} \cos \frac{x}{2})^2

[INDENT]

\displaystyle = (2 \sin ^2 \frac{x}{2})(2 \cos ^2 \frac{x}{2})

\displaystyle = (1 - \cos x)(1 + \cos x)

\displaystyle = 1 - \cos^2 x

[/INDENT]

I did quite like that for its seeming simplicity but given that it relies at line 2 to 3 on

cos2x = 1 - 2sin^2x = 2cos^2x - 1

it's pretty circular in nature.

Reply 97

IrrationalNumber

12

Original post by RichE

I did quite like that for its seeming simplicity but given that it relies at line 2 to 3 on

cos2x = 1 - 2sin^2x = 2cos^2x - 1

it's pretty circular in nature.

It was very hard to try and dodge that with that stupid integral proof I have above. But yeah, it is circular.