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tan(θ)=(m^2-m^1)/(1+m^2*m^1),m^2

解答

tan (θ) = \frac{m ^{2} - m ^{1}}{1 + m ^{2} \cdot m ^{1}}, m^{2}

解答

θ = arctan (\frac{m ^{2} - m}{m ^{3} + 1}), θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) + π, θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) + 2 π, θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) + 3 π, θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) + 4 π, θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) - π, θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) - 2 π, θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) - 3 π, θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) - 4 π

求解步骤

tan (θ) = \frac{m ^{2} - m ^{1}}{1 + m ^{2} m ^{1}}, m^{2}

化简

\frac{m ^{2} - m ^{1}}{1 + m ^{2} m ^{1}} : \frac{m ^{2} - m}{1 + m ^{3}}

\frac{m ^{2} - m ^{1}}{1 + m ^{2} m ^{1}}

使用法则

a^{1} = a

m^{1} = m

= \frac{m ^{2} - m}{1 + m ^{2} m}

1 + m^{2} m = 1 + m^{3}

1 + m^{2} m

m^{2} m = m^{3}

m^{2} m

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

m^{2} m = m^{2 + 1}

= m^{2 + 1}

数字相加:

2 + 1 = 3

= m^{3}

= 1 + m^{3}

= \frac{m ^{2} - m}{1 + m ^{3}}

使用反三角函数性质

tan (θ) = \frac{m ^{2} - m}{1 + m ^{3}}

tan (θ) = \frac{m ^{2} - m}{1 + m ^{3}}

的通解

tan (x) = a \Rightarrow x = arctan (a) + πn

θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) + πn

θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) + πn

在

m^{2}

范围内的解

θ = arctan (\frac{m ^{2} - m}{m ^{3} + 1}), θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) + π, θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) + 2 π, θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) + 3 π, θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) + 4 π, θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) - π, θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) - 2 π, θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) - 3 π, θ = arctan (\frac{m ^{2} - m}{1 + m ^{3}}) - 4 π

流行的例子

0=-1-tan^2(θ)+2tan(θ)

0 = - 1 - tan^{2} (θ) + 2 tan (θ)

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3 + 2 cos (x) = 0

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