I have to write the first function as shown in the second picture, i have tried many times now but I cant find the answer, can anyone help me with this? I would be thankful! 

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The expression `H(x) = cos((2*pi*x)/3) + cos((2*pi*(x-1))/3)+ cos((2*pi*(x-2))/3` has to be written in the form `a*cos(x*(x-x_0))`

Use the formula `cos a + cos b = 2*cos((a+b)/2)*cos((a-b)/2)`

`H(x) = cos((2*pi*x)/3) + cos((2*pi*(x-1))/3)+ cos((2*pi*(x-2))/3`

=

`2*cos(((2*pi*x)/3+(2*pi*(x-1))/3)/2)*cos(((2*pi*x)/3-(2*pi*(x-1))/3)/2)+ cos((2*pi*(x-2))/3 `

= `2*cos((2*pi*x)/3 - pi/3)*cos(pi/3)+ cos((2*pi*(x-2))/3)`

= `2*cos((2*pi*x)/3 - pi/3)*(1/2)+ cos((2*pi*(x-2))/3)`

= `cos((2*pi*x)/3 - pi/3)+ cos((2*pi*(x-2))/3)`

= `2*cos(((2*pi*x)/3 - pi/3 + (2*pi*(x-2))/3)/2)*cos(((2*pi*x)/3 - pi/3 - (2*pi*(x-2))/3)/2)`

= `2*cos ((2*pi*x)/3 - 5pi/6)*cos(-pi/6+4*pi/6)`

= `2*cos ((2*pi*x)/3 - 5pi/6)*cos(3*pi/6)`

= `2*cos ((2*pi*x)/3 - 5pi/6)*cos(pi/2)`

= `2*cos ((2*pi*x)/3 - 5pi/6)*0`

The expression `H(x) = cos((2*pi*x)/3) + cos((2*pi*(x-1))/3)+ cos((2*pi*(x-2))/3` is 0 for all values of x.

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aruv | High School Teacher | (Level 2) Valedictorian

Posted on

`H(x)=cos((2pix)/3)+cos((2pi(x-1))/3)+cos((2pi(x-2))/3))`

`=cos((2pix)/3)+cos((2pix)/3-(2pi)/3)+cos((2pix)/3-(4pi)/3)`

`since`

` cos(-a)=cos(a)`

`therefore`

`=cos((2pix)/3)+cos(pi/2+pi/6-(2pix)/3)+cos(pi+pi/3-(2pix)/3)`

`=cos((2pix)/3)-sin(pi/6-(2pix)/3)-cos(pi/3-(2pix)/3)`

`=cos((2pix)/3)-sin(pi/6)cos((2pix)/3)+sin((2pix)/3)cos(pi/6)-cos(pi/3)cos((2pix)/3)-sin(pi/3)sin((2pix)/3)`

`=cos((2pix)/3)-(1/2)cos((2pix)/3)+sqrt(3)/2sin((2pix)/3)-(1/2)cos((2pix)/3)-sqrt(3)/2sin((2pix)/3)`

`=0`

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Unknown93 | (Level 1) eNoter

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Second picture

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Unknown93 | (Level 1) eNoter

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This form

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Unknown93 | (Level 1) eNoter

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Acos(w(x-X0)) I tried to upload as a picture many times, but failed. This form I have to write the function above! 

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