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Trig identity (hard)

Prove

Unparseable latex formula:

cos(x+\frac{1}{4}n\pi ) - sin(x+\frac{1}{4}\ n\pi ) = \sqrt{2} cos(x+\frac{1}{4}(n+1)\pi ))}

(edited 9 years ago)
Original post by ubisoft
Prove

Unparseable latex formula:

cos(x+\frac{1}{4}n\pi ) - sin(x+\frac{1}{4}\ n\pi ) = \sqrt{2} cos(x+\frac{1}{4}(n+1)\pi ))}



Just use harmonic form

Cos(y) - Sin(y) = R Cos(y+a)


R will be 2\sqrt2

a will be π4\dfrac{\pi}{4}
(edited 9 years ago)
Reply 2
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ubisoft
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Original post by TenOfThem
Just use harmonic form


Haven't done that yet as I'm still on C3...

EDIT: how do you work out R and a?
Original post by ubisoft
Haven't done that yet as I'm still on C3...

EDIT: how do you work out R and a?


Ok

Start with the RHS

2cos⁡((x+n4π)+π4)\sqrt2 \cos((x+\frac{n}{4}\pi) + \frac{\pi}{4})


So Cos(A+B) where B = π4\frac{\pi}{4}
math is sexy!
Reply 5
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ubisoft
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Original post by TenOfThem
Ok

Start with the RHS

2cos⁡((x+n4π)+π4)\sqrt2 \cos((x+\frac{n}{4}\pi) + \frac{\pi}{4})


So Cos(A+B) where B = π4\frac{\pi}{4}


So then do you use the double angle properties to split it up?
Original post by ubisoft
So then do you use the double angle properties to split it up?


The compound angle rules, yes
Reply 7
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ubisoft
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3
Original post by TenOfThem
The compound angle rules, yes


Thanks, I've done it now. It was from a proof by induction question from FP2, I had to look up the harmonic form thing and it became clear.

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