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https://8dedc505ac3fba908c50836f59059ccce5cd0f1e.googledrive.com/host/0B1ZiqBksUHNYdHIxUkJmdndfMlE/June%202015%20MA%20-%20FP2%20Edexcel.pdf

hey guys its me again hehe :h::h:

can someone please tell me why in q2 (part b&c), the part inside the square brackets of the modulus-argument form equalls 1?

I have never seen this before, sooo confused :s-smilie:
Reply 1
Original post by Katiee224
https://8dedc505ac3fba908c50836f59059ccce5cd0f1e.googledrive.com/host/0B1ZiqBksUHNYdHIxUkJmdndfMlE/June%202015%20MA%20-%20FP2%20Edexcel.pdf

hey guys its me again hehe :h::h:

can someone please tell me why in q2 (part b&c), the part inside the square brackets of the modulus-argument form equalls 1?

I have never seen this before, sooo confused :s-smilie:


12π3=4π\frac{12\pi}{3} = 4\pi. cos(4π)=1\cos (4\pi) = 1 and sin(4π)=0\sin (4\pi) = 0.

So cos4π+isin4π=1+i(0)=1\cos 4\pi + i \sin 4\pi = 1 + i(0) = 1
Reply 2
Same goes for 6π3=2π\frac{6\pi}{3} = 2\pi. We have cos2π=1\cos 2\pi = 1 and sin2π=0\sin 2\pi = 0
Reply 3
Original post by Zacken
Same goes for 6π3=2π\frac{6\pi}{3} = 2\pi. We have cos2π=1\cos 2\pi = 1 and sin2π=0\sin 2\pi = 0


thank you so much!! :biggrin:
Reply 4
Original post by Katiee224
thank you so much!! :biggrin:


Makes sense? :smile:

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