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受欢迎的三角函数 >

1+tan^2(θ)=4tan^2(θ)

解答

1 + tan^{2} (θ) = 4 tan^{2} (θ)

解答

θ = 0.52359 \dots + πn, θ = - 0.52359 \dots + πn

+1

度数

θ = 3 0^{\circ} + 18 0^{\circ} n, θ = - 3 0^{\circ} + 18 0^{\circ} n

求解步骤

1 + tan^{2} (θ) = 4 tan^{2} (θ)

用替代法求解

1 + tan^{2} (θ) = 4 tan^{2} (θ)

令

: tan (θ) = u

1 + u^{2} = 4 u^{2}

1 + u^{2} = 4 u^{2} : u = \frac{1}{3}, u = - \frac{1}{3}

1 + u^{2} = 4 u^{2}

将

1

到右边

1 + u^{2} = 4 u^{2}

两边减去

1

1 + u^{2} - 1 = 4 u^{2} - 1

化简

u^{2} = 4 u^{2} - 1

u^{2} = 4 u^{2} - 1

将

4 u^{2}

para o lado esquerdo

u^{2} = 4 u^{2} - 1

两边减去

4 u^{2}

u^{2} - 4 u^{2} = 4 u^{2} - 1 - 4 u^{2}

化简

- 3 u^{2} = - 1

- 3 u^{2} = - 1

两边除以

- 3

- 3 u^{2} = - 1

两边除以

- 3

\frac{- 3 u ^{2}}{- 3} = \frac{- 1}{- 3}

化简

u^{2} = \frac{1}{3}

u^{2} = \frac{1}{3}

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = \frac{1}{3}, u = - \frac{1}{3}

u = tan (θ)

代回

tan (θ) = \frac{1}{3}, tan (θ) = - \frac{1}{3}

tan (θ) = \frac{1}{3}, tan (θ) = - \frac{1}{3}

tan (θ) = \frac{1}{3} : θ = arctan (\frac{1}{3}) + πn

tan (θ) = \frac{1}{3}

使用反三角函数性质

tan (θ) = \frac{1}{3}

tan (θ) = \frac{1}{3}

的通解

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

θ = arctan (\frac{1}{3}) + πn

θ = arctan (\frac{1}{3}) + πn

tan (θ) = - \frac{1}{3} : θ = arctan (- \frac{1}{3}) + πn

tan (θ) = - \frac{1}{3}

使用反三角函数性质

tan (θ) = - \frac{1}{3}

tan (θ) = - \frac{1}{3}

的通解

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

θ = arctan (- \frac{1}{3}) + πn

θ = arctan (- \frac{1}{3}) + πn

合并所有解

θ = arctan (\frac{1}{3}) + πn, θ = arctan (- \frac{1}{3}) + πn

以小数形式表示解

θ = 0.52359 \dots + πn, θ = - 0.52359 \dots + πn

作图

流行的例子

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(tan (θ) + 1) (cos (θ) + 1) (sin (θ)) = 0

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sin(x)= 1/(1.33)

sin (x) = \frac{1}{1.33}