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UnkleRhaukus
 one year ago
Show that:
\[\cos^2 (nx) = \frac{1 + \cos (2nx)}2\]
UnkleRhaukus
 one year ago
Show that: \[\cos^2 (nx) = \frac{1 + \cos (2nx)}2\]

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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Does 1+cos(2x) = 2cos^2(x) not work here?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I would start with \(\cos^2nx+\sin^2nx=1\)

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0\[LHS=1\sin^2nx\]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0@Astrophysics how can i show: \[1+\cos(2x) = 2\cos^2(x)\]

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.2@UnkleRhaukus use the identity for \(\cos(A+B)\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[1+\cos(2x) = \mathcal{R}(1+e^{i(2x)}) = \mathcal{R}(1+e^{ix}e^{ix}) \\~\\= \mathcal{R}(1+(\cos x+i\sin x)(\cos x+i\sin x))\\~\\ =\cdots \]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2You can use cos(A+B) = cosAsinB+sinAcosB cos(2x) = cos(x+x) = .... you see :P

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0\(\mathcal R\) is \(\Re\), the real component ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440057813193:dw

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0\[RHS= \frac{1 + \cos nx\sin nx+\sin nx\cos nx}2\\ =\frac{1 + 2\cos nx\sin nx}2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440057916809:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1That will do dw:1440058025993:dw

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.2correction @Astrophysics \[\huge\color{red}{ \cos(A+B)=\cos A\cos B\sin A\sin B}\]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0\[RHS= \frac{1 + \cos nx\cos nx\sin nx\sin nx}2\\ =\frac{1 + \cos^2nx\sin^2nx}2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440058094301:dw

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.2Now use \(\sin^2x+\cos^2x=1\)

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0\[\vdots\\=\frac{\cos^2nx+\cos^2nx}2\\ =\cos^2nx\\ =LHS\]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0got it now, not too hard (when using the right formulae)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Thanks @amilapsn I just realized the mistake lol late night...

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0[ no need to over complexify the problem ]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0theres some mistake in there i cant put my finger on it yet

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0i suppose \[\cos^2nx+\sin^2nx=1\]follows directly from \[\cos^2\theta +\sin^2\theta=1\] with \(nx=\theta\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440058675021:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1dw:1440058596216:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1try a formula for \(\cos^n(x)\) using the trig

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1then you will appreciate how complexifying really simplfies things ;p

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0even for this problem i think its still faster than just trig identities

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Yes it is cute only if you know complex numbers.. otherwise it is complex

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Nice pun ganeshie

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.2complexity to simplicity..... I like that...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1show that \[\sin(nx) = \binom{n}{1}\sin x\cos^{n1}x\binom{n}{3}\sin^3 x\cos^{n3}x+\binom{n}{5}\sin^5 x\cos^{n5}x\cdots\] where \(n\in \mathbb{Z^{+}}\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1try proving that using trig, then il show a one line proof by complexifying :)

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0if you change the question, certain methods will be more or less appropriate

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1sure, the beauty lies in the eyes of beholder :) for cos(nx) we have \[\cos(nx) = \binom{n}{0}\sin x\cos^{n}x\binom{n}{2}\sin^2 x\cos^{n2}x+\binom{n}{4}\sin^4 x\cos^{n4}x\cdots\] where \(n\in \mathbb{Z^{+}}\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1i know at least 3 different methods to prove those identities : 1) trig 2) vectors 3) complex numbers all 3 methods provide different ways to look at the problem, personally i like complex numbers because they give the answer in fewer steps compared to other methods that i am familiar with

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is the same proof as for \(\cos(x)^2=\frac12(1+\cos(2x))\) i.e. without the \(n\)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Vectors...X_X that sounds the most complicated how do you prove it with vectors?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0consider: $$\cos(2x)=\cos^2(x)\sin^2(x)\\1=\cos^2(x)+\sin^2(x)$$so it follows adding gives $$\cos(2x)+1=2\cos^2(x)\\\cos^2(x)=\frac12(1+\cos(2x))$$

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I think complex numbers are nothing but vectors in 2 dimensions with some attitude

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0i can't remember this formula \[\cos(2x)=\cos^2(x)\sin^2(x)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well, \(\cos(x+y)=\cos x\cos y\sin x\sin y\)

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0yeah, i have to derive it form the sum/difference formula \[\cos(a\pm b)=\cos a\cos b\mp \sin a\sin b\]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0But how i am supposed to remember this one \[\cos(a\pm b)=\cos a\cos b\mp \sin a\sin b\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[\cos(a\pm b)=\Re e^{i(a\pm b)}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0https://en.wikipedia.org/wiki/Proofs_of_trigonometric_identities#Cosine

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for \(\cos nx, \sin nx\) we can use recursion usign the angle sum/difference identities since \(\cos((n+1)x)=\cos(nx+x)=\dots\) and similarly in the other case
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