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

1. Astrophysics

Does 1+cos(2x) = 2cos^2(x) not work here?

2. ganeshie8

I would start with $$\cos^2nx+\sin^2nx=1$$

3. UnkleRhaukus

$LHS=1-\sin^2nx$

4. UnkleRhaukus

@Astrophysics how can i show: $1+\cos(2x) = 2\cos^2(x)$

5. amilapsn

@UnkleRhaukus use the identity for $$\cos(A+B)$$

6. ganeshie8

$1+\cos(2x) = \mathcal{R}(1+e^{i(2x)}) = \mathcal{R}(1+e^{ix}e^{ix}) \\~\\= \mathcal{R}(1+(\cos x+i\sin x)(\cos x+i\sin x))\\~\\ =\cdots$

7. Astrophysics

You can use cos(A+B) = cosAsinB+sinAcosB cos(2x) = cos(x+x) = .... you see :P

8. UnkleRhaukus

$$\mathcal R$$ is $$\Re$$, the real component ?

9. ganeshie8

Yes

10. anonymous

|dw:1440057813193:dw|

11. UnkleRhaukus

$RHS= \frac{1 + \cos nx\sin nx+\sin nx\cos nx}2\\ =\frac{1 + 2\cos nx\sin nx}2$

12. anonymous

|dw:1440057916809:dw|

13. ganeshie8

That will do |dw:1440058025993:dw|

14. amilapsn

correction @Astrophysics $\huge\color{red}{ \cos(A+B)=\cos A\cos B-\sin A\sin B}$

15. UnkleRhaukus

$RHS= \frac{1 + \cos nx\cos nx-\sin nx\sin nx}2\\ =\frac{1 + \cos^2nx-\sin^2nx}2$

16. anonymous

|dw:1440058094301:dw|

17. amilapsn

Now use $$\sin^2x+\cos^2x=1$$

18. UnkleRhaukus

$\vdots\\=\frac{\cos^2nx+\cos^2nx}2\\ =\cos^2nx\\ =LHS$

19. UnkleRhaukus

got it now, not too hard (when using the right formulae)

20. Astrophysics

Thanks @amilapsn I just realized the mistake lol late night...

21. amilapsn

Yw! lol :D

22. UnkleRhaukus

[ no need to over complexify the problem ]

23. anonymous

something is wrong

24. anonymous

theres some mistake in there i cant put my finger on it yet

25. UnkleRhaukus

i suppose $\cos^2nx+\sin^2nx=1$follows directly from $\cos^2\theta +\sin^2\theta=1$ with $$nx=\theta$$

26. anonymous

ohh i got it xD

27. anonymous

|dw:1440058675021:dw|

28. ganeshie8

|dw:1440058596216:dw|

29. ganeshie8

try a formula for $$\cos^n(x)$$ using the trig

30. ganeshie8

then you will appreciate how complexifying really simplfies things ;p

31. anonymous

even for this problem i think its still faster than just trig identities

32. amilapsn

lol...

33. ganeshie8

Yes it is cute only if you know complex numbers.. otherwise it is complex

34. Astrophysics

Nice pun ganeshie

35. amilapsn

complexity to simplicity..... I like that...

36. 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 $$n\in \mathbb{Z^{+}}$$

37. ganeshie8

try proving that using trig, then il show a one line proof by complexifying :)

38. UnkleRhaukus

if you change the question, certain methods will be more or less appropriate

39. 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 $$n\in \mathbb{Z^{+}}$$

40. 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

41. anonymous

this is the same proof as for $$\cos(x)^2=\frac12(1+\cos(2x))$$ i.e. without the $$n$$

42. Astrophysics

Vectors...X_X that sounds the most complicated how do you prove it with vectors?

43. anonymous

consider: $$\cos(2x)=\cos^2(x)-\sin^2(x)\\1=\cos^2(x)+\sin^2(x)$$so it follows adding gives $$\cos(2x)+1=2\cos^2(x)\\\cos^2(x)=\frac12(1+\cos(2x))$$

44. ganeshie8

I think complex numbers are nothing but vectors in 2 dimensions with some attitude

45. UnkleRhaukus

i can't remember this formula $\cos(2x)=\cos^2(x)-\sin^2(x)$

46. anonymous

well, $$\cos(x+y)=\cos x\cos y-\sin x\sin y$$

47. UnkleRhaukus

yeah, i have to derive it form the sum/difference formula $\cos(a\pm b)=\cos a\cos b\mp \sin a\sin b$

48. UnkleRhaukus

But how i am supposed to remember this one $\cos(a\pm b)=\cos a\cos b\mp \sin a\sin b$

49. ganeshie8

$\cos(a\pm b)=\Re e^{i(a\pm b)}$

50. anonymous
51. UnkleRhaukus

hmm.

52. 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