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受欢迎的三角函数 >

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

作图

流行的例子

tan^2(x)=3-2tan(x)

tan^{2} (x) = 3 - 2 tan (x)

tan(3x)+tan(5x)=0

tan (3 x) + tan (5 x) = 0

sin(x)= 1/2 sqrt(3)

sin (x) = \frac{1}{2} 3

2 cot (x) + 4 = 6

sin(θ)=(20N)/(35N)

sin (θ) = \frac{20 N}{35 N}