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I want to rewrite the following expression into arctan.

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safisay
safisay 2017 年 11 月 30 日
回答済み: Walter Roberson 2017 年 11 月 30 日
Hi All,
I want to rewrite the following expression into arctan. Can anyone help?
m=(k_1*(sin(theta1) + cos(k_1*(gama - theta1))*sin(theta0) - cos(theta0)*sin(theta1) + cos(theta1)*sin(theta0) - M*sin(theta1) - k_1*sin(k_1*(gama - theta1)) - M*cos(k_1*(gama - theta1))*sin(theta1) + k_1*cos(theta0)*sin(k_1*(gama - theta1)) + M*k_1*sin(k_1*(gama - theta1)) - M*k_1*cos(theta1)*sin(k_1*(gama - theta1))))/(M*l*(2*k_1 - sin(k_1*(gama - theta1))*sin(theta1) + 2*k_1*cos(k_1*(gama - theta1))*cos(theta1) - k_1^2*sin(k_1*(gama - theta1))*sin(theta1))) - (k_1*(M*sin(theta1) - cos(k_1*(gama - theta1))*sin(theta1) - sin(theta0) + M*cos(k_1*(gama - theta1))*sin(theta1) - k_1*cos(theta1)*sin(k_1*(gama - theta1)) + cos(k_1*(gama - theta1))*cos(theta0)*sin(theta1) - cos(k_1*(gama - theta1))*cos(theta1)*sin(theta0) - M*k_1*sin(k_1*(gama - theta1)) + k_1*cos(theta0)*cos(theta1)*sin(k_1*(gama - theta1)) + k_1*sin(k_1*(gama - theta1))*sin(theta0)*sin(theta1) + M*k_1*cos(theta1)*sin(k_1*(gama - theta1))))/(M*(2*k_1 - sin(k_1*(gama - theta1))*sin(theta1) + 2*k_1*cos(k_1*(gama - theta1))*cos(theta1) - k_1^2*sin(k_1*(gama - theta1))*sin(theta1)))

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Walter Roberson
Walter Roberson 2017 年 11 月 30 日
No, there are no log terms and no imaginary constants and no sqrt() of expressions; there is nothing to rewrite into arctan. It would make more sense to rewrite it into tan
Some of the trig calls can be combined, for whatever that is worth. That does not make it any easier to understand, and has nothing to do with rewriting to arctan.

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