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標題:
[Maths] Anyone can show me how to do this question ,plz~
發問:
a.) Find the values of the following sums.cos(0)+cos(2π/2) cos(0)+cos(2π/3)+cos(2 2π/3) cos(0)+cos(2π/4)+cos(2 2π/4)+cos(3 2π/4)cos(0)+cos(2π/5)+cos(2 2π/5)+cos(3 2π/5)+cos (4 2π/5)b.) Base on (a.) what would you conjecture about the... 顯示更多 a.) Find the values of the following sums. cos(0)+cos(2π/2) cos(0)+cos(2π/3)+cos(2 2π/3) cos(0)+cos(2π/4)+cos(2 2π/4)+cos(3 2π/4) cos(0)+cos(2π/5)+cos(2 2π/5)+cos(3 2π/5)+cos (4 2π/5) b.) Base on (a.) what would you conjecture about the sum C= cos(0)+cos(x)+cos(2x)+....+cos((n-1)x) when x= 2π/n Here n is a positive integer c.) Simplify sin(A+B)-sin(A-B) and sin(kx+x/2)-sin(kx-x/2) Use your result to get a simpler form of C= cos(0)+cos(x)+cos(2x)+....+cos((n-1)x) for general x d.) Can you use the result in (c.) to prove your conjecture from (b.) e.) Useing these ideas can you calculate the values of S= sin(0)+sin(x)+sin(2x)+....+sin((n-1)x) when x=2π/n T= sin(A)+sin(A+2π/n)+sin(A+4π/n)+ ...+sin(A+2π(n-1)/n) here A is a general angle f.) What is the esstential reason that these calculations work? 更新: cos(2 2π/3) <---= cos(2 and 2π/3)
最佳解答:
a.) Find the values of the following sums. cos(0)+cos(2π/2) = 1-1 = 0 cos(0)+cos(2π/3)+cos(2 2π/3) = 1-0.5-0.5 = 0 cos(0)+cos(2π/4)+cos(2 2π/4)+cos(3 2π/4) = 1+0-1+0 = 0 cos(0)+cos(2π/5)+cos(2 2π/5)+cos(3 2π/5)+cos (4 2π/5) = 1+0.309-0.809-0.809+0.309 = 0 b.) Base on (a.) what would you conjecture about the sum C= cos(0)+cos(x)+cos(2x)+....+cos((n-1)x) when x= 2π/n Here n is a positive integer C= cos(0)+cos(x)+cos(2x)+....+cos((n-1)x) when x= 2π/n = 0 c.) Simplify sin(A+B)-sin(A-B) and sin(kx+x/2)-sin(kx-x/2) sin(A+B)-sin(A-B) = 2sinBcosA and sin(kx+x/2)-sin(kx-x/2) = 2sin(x/2)cos(kx) Use your result to get a simpler form of C= cos(0)+cos(x)+cos(2x)+....+cos((n-1)x) for general x 2Csin(x/2) = 2sin(x/2)[cos(0)+cos(x)+cos(2x)+....+cos((n-1)x)] = (sin(x/2)-sin(-x/2)) + (sin(x+x/2)-sin(x-x/2)) + ... + (sin((n-1)x+x/2) -sin ((n-1)x-x/2)) = sin(nx-x/2)-sin(-x/2) = sin(nx-x/2)+sin(x/2) C = 1/2[sin(nx-x/2)/sin(x/2) + 1] d.) Can you use the result in (c.) to prove your conjecture from (b.) C= cos(0)+cos(x)+cos(2x)+....+cos((n-1)x) when x= 2π/n Here n is a positive integer C = 1/2[sin(nx-x/2)/sin(x/2) + 1] C = 1/2[sin(2π- π/n)/sin(π/n) + 1] C = 1/2[-sin(π/n)/sin(π/n) + 1] C = 1/2[-1+1] C = 0
其他解答:
[Maths] Anyone can show me how to do this question ,plz~
發問:
a.) Find the values of the following sums.cos(0)+cos(2π/2) cos(0)+cos(2π/3)+cos(2 2π/3) cos(0)+cos(2π/4)+cos(2 2π/4)+cos(3 2π/4)cos(0)+cos(2π/5)+cos(2 2π/5)+cos(3 2π/5)+cos (4 2π/5)b.) Base on (a.) what would you conjecture about the... 顯示更多 a.) Find the values of the following sums. cos(0)+cos(2π/2) cos(0)+cos(2π/3)+cos(2 2π/3) cos(0)+cos(2π/4)+cos(2 2π/4)+cos(3 2π/4) cos(0)+cos(2π/5)+cos(2 2π/5)+cos(3 2π/5)+cos (4 2π/5) b.) Base on (a.) what would you conjecture about the sum C= cos(0)+cos(x)+cos(2x)+....+cos((n-1)x) when x= 2π/n Here n is a positive integer c.) Simplify sin(A+B)-sin(A-B) and sin(kx+x/2)-sin(kx-x/2) Use your result to get a simpler form of C= cos(0)+cos(x)+cos(2x)+....+cos((n-1)x) for general x d.) Can you use the result in (c.) to prove your conjecture from (b.) e.) Useing these ideas can you calculate the values of S= sin(0)+sin(x)+sin(2x)+....+sin((n-1)x) when x=2π/n T= sin(A)+sin(A+2π/n)+sin(A+4π/n)+ ...+sin(A+2π(n-1)/n) here A is a general angle f.) What is the esstential reason that these calculations work? 更新: cos(2 2π/3) <---= cos(2 and 2π/3)
最佳解答:
a.) Find the values of the following sums. cos(0)+cos(2π/2) = 1-1 = 0 cos(0)+cos(2π/3)+cos(2 2π/3) = 1-0.5-0.5 = 0 cos(0)+cos(2π/4)+cos(2 2π/4)+cos(3 2π/4) = 1+0-1+0 = 0 cos(0)+cos(2π/5)+cos(2 2π/5)+cos(3 2π/5)+cos (4 2π/5) = 1+0.309-0.809-0.809+0.309 = 0 b.) Base on (a.) what would you conjecture about the sum C= cos(0)+cos(x)+cos(2x)+....+cos((n-1)x) when x= 2π/n Here n is a positive integer C= cos(0)+cos(x)+cos(2x)+....+cos((n-1)x) when x= 2π/n = 0 c.) Simplify sin(A+B)-sin(A-B) and sin(kx+x/2)-sin(kx-x/2) sin(A+B)-sin(A-B) = 2sinBcosA and sin(kx+x/2)-sin(kx-x/2) = 2sin(x/2)cos(kx) Use your result to get a simpler form of C= cos(0)+cos(x)+cos(2x)+....+cos((n-1)x) for general x 2Csin(x/2) = 2sin(x/2)[cos(0)+cos(x)+cos(2x)+....+cos((n-1)x)] = (sin(x/2)-sin(-x/2)) + (sin(x+x/2)-sin(x-x/2)) + ... + (sin((n-1)x+x/2) -sin ((n-1)x-x/2)) = sin(nx-x/2)-sin(-x/2) = sin(nx-x/2)+sin(x/2) C = 1/2[sin(nx-x/2)/sin(x/2) + 1] d.) Can you use the result in (c.) to prove your conjecture from (b.) C= cos(0)+cos(x)+cos(2x)+....+cos((n-1)x) when x= 2π/n Here n is a positive integer C = 1/2[sin(nx-x/2)/sin(x/2) + 1] C = 1/2[sin(2π- π/n)/sin(π/n) + 1] C = 1/2[-sin(π/n)/sin(π/n) + 1] C = 1/2[-1+1] C = 0
其他解答:
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