Вычислите значение примера
(sina/cos2a+cosa/sin2a)*((sina+sin7a)/cosa) если а=п/9
Упростите пример
((cos^2a+4cos^2a/2sin^2a/2)(1-2sin^2a/2))/cosa+cos3a
(sina/cos2a+cosa/sin2a)*((sina+sin7a)/cosa) если а=п/9
Упростите пример
((cos^2a+4cos^2a/2sin^2a/2)(1-2sin^2a/2))/cosa+cos3a
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Answers & Comments
Verified answer
A=(sinα*sin2α+cosαα*cos2α)/(sin2α*cos2α)*(2sin4α*cos3α)/cosα== cos(2α - α)/(1/2sin4α)*(2sin4α*cos3α)/cosα= 4cosα*cos3α/cosα=4cos3α;
при = π/9 выражения принимает значение :
A= 4cos3*π/9=4cosππ/3=4*1/2 =2.
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((cos^2a+4cos^2a/2sin^2a/2)(1-2sin^2a/2))/cosa+cos3a
= (cos²α +(2cosα/2*sinα/2)²)*cosα/cos +cos3α=
== (cos²α +sinα²)+cos3α =1+cos3α = 2cos²(3α/2)
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