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prove that |1 - cis(theta)| = 2sin1/2(theta) .... and vector harddd question!

1)prove that |1 - cis(theta)| = 2sin1/2(theta) for 0</= theta </= 2pi.... where cis(theta) is cos theta + i sin theta.

2) Let A be l*m and B be m*n and u be an n-component vector. Write down the j-th component of Bu and the i-th component of A(Bu) using summation notation. Write down also the i,k component of AB and the i-th component of (AB)u and deduce that A(Bu) = (AB)u

Reply 1

|1-cosx - i sinx| = rt((1-cosx)^2 + (sinx)^2) = rt(1 - 2cosx + cos^2 x + sin^2 x) = rt(2-2cosx)

now, cosx = 1 - 2sin^2 x/2
sin x/2 = rt((1-cosx)/2)
so 2sin x/2 = rt2rt(1-cosx)

Reply 2

OP

thanks for that.... how on earth do i do the next question.. rep will be given.,

Reply 3

See attachment.

matrices2.png79.9KB

Reply 4

sul

1)prove that |1 - cis(theta)| = 2sin1/2(theta) for 0</= theta </= 2pi.... where cis(theta) is cos theta + i sin theta.

|1-\cos(\theta)-i\sin\theta|=|2\sin^2(\frac{\theta}{2})-2i\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})|=|-2i||\sin(\frac{\theta}{2})||\cos(\frac{\theta}{2})+i\sin(\frac{\theta}{2})|=2\sin(\frac{\theta}{2})

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