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Nov 19, 2016 · 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 ...

Oct 9, 2017 · 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^@)#

Oct 5, 2017 · #cos20^@cos40^@cos80^@,# #=1/(2sin20^@)(2sin20^@cos20^@)cos40^@cos80^@,# …

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

Oct 8, 2017 · 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

Feb 2, 2025 · $\begingroup$ $\cos20^{\circ}$ 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 …

Sep 27, 2015 · 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