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 ...

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Value of $\cos 20^\circ + 2 \sin^2 55^\circ - \sqrt{2} \sin 65^\circ$. Frankly speaking, I have no idea how to start in this question.

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2月28日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

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

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

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?

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 …

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 }#?