Show that:

$${\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{x}}\right)}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{4}}}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{x}}\right)} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{x}}\right)}{\mathtt{\,\times\,}}{{cos}\left(}^{{\mathtt{2}}}{\mathtt{\,\times\,}}\left({\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{x}}\right)$$

I got that πx can be replaced by a θ: That helped me for a little bit, but then I got stuck. My current place is as stated:

$${\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}\left({\mathtt{2}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{x}}\right)}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{x}}\right)}\right){\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{4}}}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}\right)} = {\mathtt{2}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{x}}\right)}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{x}}\right)}{\mathtt{\,\times\,}}{{cos}\left(}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\mathtt{x}}$$

How should I go about finding how the two equations are equivalent from here?

Thanks

-GoldenLeaf

GoldenLeaf
Apr 6, 2015

#1**+15 **

(1/2) sin(2θ) + (1/4)sin(4θ) =

(1/2)*2sinθcosθ + (1/4)sin(2θ + 2θ) =

sinθcosθ + (1/4)[ sin2θcos2θ + sin2θcos2θ ] =

sinθcosθ + (1/4) [ (2)[sin2θ][cos2θ] ] =

sinθcosθ + (1/4) (2) [ [sin2θ][cos2θ] ] =

sinθcosθ + (1/2)[2sinθcosθ]* [2cos^2θ - 1] =

sinθcosθ + sinθcosθ * [ 2cos^2θ - 1 ] =

sinθcosθ + sinθcosθ * 2cos^2θ - sinθcosθ =

sinθcosθ * 2cos^2θ =

2 sinθcosθ * cos^2θ

sin(2θ) * cos^2θ and replacing θ with pi * x .....we have

sin(2* pi *x ) * cos^2(pi * x)

And that's it, GL ....!!!!

BTW......here's a graph.......https://www.desmos.com/calculator/cd2jja1kws

As GL has stated...it is indeed "saw-toothed" ......very odd !!!!

CPhill
Apr 6, 2015

#1**+15 **

Best Answer

(1/2) sin(2θ) + (1/4)sin(4θ) =

(1/2)*2sinθcosθ + (1/4)sin(2θ + 2θ) =

sinθcosθ + (1/4)[ sin2θcos2θ + sin2θcos2θ ] =

sinθcosθ + (1/4) [ (2)[sin2θ][cos2θ] ] =

sinθcosθ + (1/4) (2) [ [sin2θ][cos2θ] ] =

sinθcosθ + (1/2)[2sinθcosθ]* [2cos^2θ - 1] =

sinθcosθ + sinθcosθ * [ 2cos^2θ - 1 ] =

sinθcosθ + sinθcosθ * 2cos^2θ - sinθcosθ =

sinθcosθ * 2cos^2θ =

2 sinθcosθ * cos^2θ

sin(2θ) * cos^2θ and replacing θ with pi * x .....we have

sin(2* pi *x ) * cos^2(pi * x)

And that's it, GL ....!!!!

BTW......here's a graph.......https://www.desmos.com/calculator/cd2jja1kws

As GL has stated...it is indeed "saw-toothed" ......very odd !!!!

CPhill
Apr 6, 2015