+0  
 
+2
1147
3
avatar+1006 

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

 Apr 6, 2015

Best Answer 

 #1
avatar+121064 
+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  !!!!

 

  

 Apr 6, 2015
 #1
avatar+121064 
+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
 #2
avatar+1006 
+5

BLESS YOU CPHILL ALL MY LOVE TO YOU

 Apr 6, 2015
 #3
avatar+115412 
+10

Hi GoledenLeaf, great work Chris 

I love the way formulas can make graphs that have such weird shapes.

3D graphs get even better.   

 Apr 6, 2015

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