UnkleRhaukus
  • UnkleRhaukus
Show that: \[\cos^2 (nx) = \frac{1 + \cos (2nx)}2\]
Trigonometry
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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Astrophysics
  • Astrophysics
Does 1+cos(2x) = 2cos^2(x) not work here?
ganeshie8
  • ganeshie8
I would start with \(\cos^2nx+\sin^2nx=1\)
UnkleRhaukus
  • UnkleRhaukus
\[LHS=1-\sin^2nx\]

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More answers

UnkleRhaukus
  • UnkleRhaukus
@Astrophysics how can i show: \[1+\cos(2x) = 2\cos^2(x)\]
amilapsn
  • amilapsn
@UnkleRhaukus use the identity for \(\cos(A+B)\)
ganeshie8
  • ganeshie8
\[1+\cos(2x) = \mathcal{R}(1+e^{i(2x)}) = \mathcal{R}(1+e^{ix}e^{ix}) \\~\\= \mathcal{R}(1+(\cos x+i\sin x)(\cos x+i\sin x))\\~\\ =\cdots \]
Astrophysics
  • Astrophysics
You can use cos(A+B) = cosAsinB+sinAcosB cos(2x) = cos(x+x) = .... you see :P
UnkleRhaukus
  • UnkleRhaukus
\(\mathcal R\) is \(\Re\), the real component ?
ganeshie8
  • ganeshie8
Yes
anonymous
  • anonymous
|dw:1440057813193:dw|
UnkleRhaukus
  • UnkleRhaukus
\[RHS= \frac{1 + \cos nx\sin nx+\sin nx\cos nx}2\\ =\frac{1 + 2\cos nx\sin nx}2\]
anonymous
  • anonymous
|dw:1440057916809:dw|
ganeshie8
  • ganeshie8
That will do |dw:1440058025993:dw|
amilapsn
  • amilapsn
correction @Astrophysics \[\huge\color{red}{ \cos(A+B)=\cos A\cos B-\sin A\sin B}\]
UnkleRhaukus
  • UnkleRhaukus
\[RHS= \frac{1 + \cos nx\cos nx-\sin nx\sin nx}2\\ =\frac{1 + \cos^2nx-\sin^2nx}2\]
anonymous
  • anonymous
|dw:1440058094301:dw|
amilapsn
  • amilapsn
Now use \(\sin^2x+\cos^2x=1\)
UnkleRhaukus
  • UnkleRhaukus
\[\vdots\\=\frac{\cos^2nx+\cos^2nx}2\\ =\cos^2nx\\ =LHS\]
UnkleRhaukus
  • UnkleRhaukus
got it now, not too hard (when using the right formulae)
Astrophysics
  • Astrophysics
Thanks @amilapsn I just realized the mistake lol late night...
amilapsn
  • amilapsn
Yw! lol :D
UnkleRhaukus
  • UnkleRhaukus
[ no need to over complexify the problem ]
anonymous
  • anonymous
something is wrong
anonymous
  • anonymous
theres some mistake in there i cant put my finger on it yet
UnkleRhaukus
  • UnkleRhaukus
i suppose \[\cos^2nx+\sin^2nx=1\]follows directly from \[\cos^2\theta +\sin^2\theta=1\] with \(nx=\theta\)
anonymous
  • anonymous
ohh i got it xD
anonymous
  • anonymous
|dw:1440058675021:dw|
ganeshie8
  • ganeshie8
|dw:1440058596216:dw|
ganeshie8
  • ganeshie8
try a formula for \(\cos^n(x)\) using the trig
ganeshie8
  • ganeshie8
then you will appreciate how complexifying really simplfies things ;p
anonymous
  • anonymous
even for this problem i think its still faster than just trig identities
amilapsn
  • amilapsn
lol...
ganeshie8
  • ganeshie8
Yes it is cute only if you know complex numbers.. otherwise it is complex
Astrophysics
  • Astrophysics
Nice pun ganeshie
amilapsn
  • amilapsn
complexity to simplicity..... I like that...
ganeshie8
  • ganeshie8
show that \[\sin(nx) = \binom{n}{1}\sin x\cos^{n-1}x-\binom{n}{3}\sin^3 x\cos^{n-3}x+\binom{n}{5}\sin^5 x\cos^{n-5}x-\cdots\] where \(n\in \mathbb{Z^{+}}\)
ganeshie8
  • ganeshie8
try proving that using trig, then il show a one line proof by complexifying :)
UnkleRhaukus
  • UnkleRhaukus
if you change the question, certain methods will be more or less appropriate
ganeshie8
  • ganeshie8
sure, the beauty lies in the eyes of beholder :) for cos(nx) we have \[\cos(nx) = \binom{n}{0}\sin x\cos^{n}x-\binom{n}{2}\sin^2 x\cos^{n-2}x+\binom{n}{4}\sin^4 x\cos^{n-4}x-\cdots\] where \(n\in \mathbb{Z^{+}}\)
ganeshie8
  • ganeshie8
i 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
  • anonymous
this is the same proof as for \(\cos(x)^2=\frac12(1+\cos(2x))\) i.e. without the \(n\)
Astrophysics
  • Astrophysics
Vectors...X_X that sounds the most complicated how do you prove it with vectors?
anonymous
  • anonymous
consider: $$\cos(2x)=\cos^2(x)-\sin^2(x)\\1=\cos^2(x)+\sin^2(x)$$so it follows adding gives $$\cos(2x)+1=2\cos^2(x)\\\cos^2(x)=\frac12(1+\cos(2x))$$
ganeshie8
  • ganeshie8
I think complex numbers are nothing but vectors in 2 dimensions with some attitude
UnkleRhaukus
  • UnkleRhaukus
i can't remember this formula \[\cos(2x)=\cos^2(x)-\sin^2(x)\]
anonymous
  • anonymous
well, \(\cos(x+y)=\cos x\cos y-\sin x\sin y\)
UnkleRhaukus
  • UnkleRhaukus
yeah, i have to derive it form the sum/difference formula \[\cos(a\pm b)=\cos a\cos b\mp \sin a\sin b\]
UnkleRhaukus
  • UnkleRhaukus
But how i am supposed to remember this one \[\cos(a\pm b)=\cos a\cos b\mp \sin a\sin b\]
ganeshie8
  • ganeshie8
\[\cos(a\pm b)=\Re e^{i(a\pm b)}\]
anonymous
  • anonymous
https://en.wikipedia.org/wiki/Proofs_of_trigonometric_identities#Cosine
UnkleRhaukus
  • UnkleRhaukus
hmm.
anonymous
  • anonymous
for \(\cos nx, \sin nx\) 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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