Bulbagarden Forums - Comments - Blogs

• |
Originally Posted by Croag
Not gonna look at the link or I'll get put off

let theta = a so it's easier
sin2a = cos2a
sin2a/cos2a = 1
tan2a = 1
a = 1/2arctan(1)
a = 1/2*π/4
a = π/8
Therefore they intersect when a = π/8

Find the first area between 0 and π/8
1/2∫[0,π/8] cos(2a) da
= 1/2[1/2sin2a]|0,π/8
= 1/4[sin(π/4)-sin(0)]
= 1/4[1/√2]
= 1/4√2

Do you know what the graph of cos2a and sin2a look like? They're like two loops about the origin but slightly rotated from each other. What I found above was a small quarter of the area you want, you can just times it by four and you get 1/√2 or √2/2 which is what that yahoo answer apparently got.

I can't possibly see how it could be π/8 - 1/4 since what you are integrating doesn't have any constants, and therefore would not give you any theta to sub in values of pi from. Using the double angle formulae is unnecessary since integrating cos^2(a) or sin^2(a) (which would give you the constant that you need) is impossible without converting it back to cos2a/sin2a.

Sorry I couldn't help more, but for all purposes since this is a simple problem, getting what the text book says is pretty impossible, and unless I'm not seeing it either, the book is maybe wrong. x)

thanks, That's what I was thinking too, but I needed an outside source, fresh on the problem to confirm it. Where's the +1 Karma button on BMGf, again?
• |
I just saw math and backed into a corner, sobbing.

But of course Croag would know what to do. :P
• |
Not gonna look at the link or I'll get put off

let theta = a so it's easier
sin2a = cos2a
sin2a/cos2a = 1
tan2a = 1
a = 1/2arctan(1)
a = 1/2*π/4
a = π/8
Therefore they intersect when a = π/8

Find the first area between 0 and π/8
1/2∫[0,π/8] cos(2a) da
= 1/2[1/2sin2a]|0,π/8
= 1/4[sin(π/4)-sin(0)]
= 1/4[1/√2]
= 1/4√2

Do you know what the graph of cos2a and sin2a look like? They're like two loops about the origin but slightly rotated from each other. What I found above was a small quarter of the area you want, you can just times it by four and you get 1/√2 or √2/2 which is what that yahoo answer apparently got.

I can't possibly see how it could be π/8 - 1/4 since what you are integrating doesn't have any constants, and therefore would not give you any theta to sub in values of pi from. Using the double angle formulae is unnecessary since integrating cos^2(a) or sin^2(a) (which would give you the constant that you need) is impossible without converting it back to cos2a/sin2a.

Sorry I couldn't help more, but for all purposes since this is a simple problem, getting what the text book says is pretty impossible, and unless I'm not seeing it either, the book is maybe wrong. x)

• |