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2016年11月19日 · The answer is =125/2(1+isqrt3) DeMoivre 'theorem is (costheta+isintheta)^n= cosntheta+isinntheta Therefore, (5(cos20+isin20))^3 =5^3(cos3*20+isin3*20) =125 (cos60 ...

2017年10月9日 · As #sinA+sinB=2sin((A+B)/2)cos((A-B)/2)#. and #cosA+cosB=2cos((A+B)/2)cos((A-B)/2)#. Hence #(sin10^@+sin20^@)/(cos10^@+cos20^@)#

2017年10月5日 · #cos20^@cos40^@cos80^@,# #=1/(2sin20^@)(2sin20^@cos20^@)cos40^@cos80^@,# …

2017年7月16日 · How do you prove that #\frac { \cos 20^ { \circ } - \sin 20^ { \circ } } { \cos 20^ { \circ } + \sin 20^ { \circ } } = \tan 25^ { \circ }#?

2017年10月8日 · 13613 views around the world You can reuse this answer ...

You would need an expression to work with. For example: Given #sinalpha=3/5# and #cosalpha=-4/5#, you could find #sin2 alpha# by using the double angle identity

2025年2月2日 · $\begingroup$ $\cos20^{\circ}$ is a closed form. But it can be proved that, to express it in radicals, nonreal numbers are needed. The usual proof goes via Galois Theory, and won't make much sense to you if you haven't studied that …

2015年9月27日 · How do you use the Squeeze Theorem to find #lim (x^2)(cos20(pi*x)) # as x approaches zero?

Question: Given the continues sinusoidal signal x(t)=cos20 mit, which is sampled with the sampling rate fs =4B. Find the number of samples that will be taken during the one period. a. 6 b. 4 c. 8 d. 2