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受欢迎的三角函数 >

sin(θ)=sqrt(1-(1-(cos(θ)-1)^2))

解答

sin (θ) = 1 - (1 - (cos (θ) - 1)^{2})

解答

θ = \frac{π}{2} + 2 πn, θ = 2 πn

+1

度数

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 0^{\circ} + 36 0^{\circ} n

求解步骤

sin (θ) = 1 - (1 - (cos (θ) - 1)^{2})

两边进行平方

sin^{2} (θ) = (1 - (1 - (cos (θ) - 1)^{2}))^{2}

两边减去

1 - (1 - (cos (θ) - 1)^{2})^{2}

sin^{2} (θ) - 1 + 2 cos (θ) - cos^{2} (θ) = 0

使用三角恒等式改写

- 1 - cos^{2} (θ) + sin^{2} (θ) + 2 cos (θ)

使用毕达哥拉斯恒等式:

1 = cos^{2} (x) + sin^{2} (x)

1 - sin^{2} (x) = cos^{2} (x)

= - cos^{2} (θ) + 2 cos (θ) - cos^{2} (θ)

化简

= - 2 cos^{2} (θ) + 2 cos (θ)

2 cos (θ) - 2 cos^{2} (θ) = 0

用替代法求解

2 cos (θ) - 2 cos^{2} (θ) = 0

令

: cos (θ) = u

2 u - 2 u^{2} = 0

2 u - 2 u^{2} = 0 : u = 0, u = 1

2 u - 2 u^{2} = 0

改写成标准形式

a x^{2} + b x + c = 0

- 2 u^{2} + 2 u = 0

使用求根公式求解

- 2 u^{2} + 2 u = 0

二次方程求根公式:

若

a = - 2, b = 2, c = 0

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

2^{2} - 4 (- 2) \cdot 0 = 2

2^{2} - 4 (- 2) \cdot 0

使用法则

- (- a) = a

= 2^{2} + 4 \cdot 2 \cdot 0

使用法则

0 \cdot a = 0

= 2^{2} + 0

2^{2} + 0 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 2

u_{1, 2} = \frac{- 2 \pm 2}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- 2 + 2}{2 ( - 2 )}, u_{2} = \frac{- 2 - 2}{2 ( - 2 )}

u = \frac{- 2 + 2}{2 ( - 2 )} : 0

\frac{- 2 + 2}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 2 + 2}{- 2 \cdot 2}

数字相加/相减:

- 2 + 2 = 0

= \frac{0}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{0}{- 4}

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{0}{4}

使用法则

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

u = \frac{- 2 - 2}{2 ( - 2 )} : 1

\frac{- 2 - 2}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 2 - 2}{- 2 \cdot 2}

数字相减:

- 2 - 2 = - 4

= \frac{- 4}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 4}{- 4}

使用分式法则:

\frac{- a}{- b} = \frac{a}{b}

= \frac{4}{4}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

u = 0, u = 1

u = cos (θ)

代回

cos (θ) = 0, cos (θ) = 1

cos (θ) = 0, cos (θ) = 1

cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

cos (θ) = 0

cos (θ) = 0

的通解

cos (x)

周期表（周期为

2 πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

cos (θ) = 1 : θ = 2 πn

cos (θ) = 1

cos (θ) = 1

的通解

cos (x)

周期表（周期为

2 πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = 0 + 2 πn

θ = 0 + 2 πn

解

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

合并所有解

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = 2 πn

将解代入原方程进行验证

将它们代入

sin (θ) = 1 - (1 - (cos (θ) - 1)^{2})

检验解是否符合

去除与方程不符的解。

检验

\frac{π}{2} + 2 πn

的解:

真

\frac{π}{2} + 2 πn

代入

n = 1

\frac{π}{2} + 2 π 1

对于

sin (θ) = 1 - (1 - (cos (θ) - 1)^{2})

代入

θ = \frac{π}{2} + 2 π 1

sin (\frac{π}{2} + 2 π 1) = 1 - (1 - (cos (\frac{π}{2} + 2 π 1) - 1)^{2})

整理后得

1 = 1

\Rightarrow 真

检验

\frac{3 π}{2} + 2 πn

的解:

假

\frac{3 π}{2} + 2 πn

代入

n = 1

\frac{3 π}{2} + 2 π 1

对于

sin (θ) = 1 - (1 - (cos (θ) - 1)^{2})

代入

θ = \frac{3 π}{2} + 2 π 1

sin (\frac{3 π}{2} + 2 π 1) = 1 - (1 - (cos (\frac{3 π}{2} + 2 π 1) - 1)^{2})

整理后得

- 1 = 1

\Rightarrow 假

检验

2 πn

的解:

真

2 πn

代入

n = 1

2 π 1

对于

sin (θ) = 1 - (1 - (cos (θ) - 1)^{2})

代入

θ = 2 π 1

sin (2 π 1) = 1 - (1 - (cos (2 π 1) - 1)^{2})

整理后得

0 = 0

\Rightarrow 真

θ = \frac{π}{2} + 2 πn, θ = 2 πn

作图

流行的例子

14.79m=((7.61_{0}^2sin(2θ_{0}))/g)

14.79 m = (\frac{7.6 1 _{0}^{2} sin ( 2 θ _{0} )}{g})

2cos(x)+3cos(x)-2=0

2 cos (x) + 3 cos (x) - 2 = 0

sec((5θ)/4)=2,[0.2pi]

sec (\frac{5 θ}{4}) = 2, [0.2 π]

cos(x)+|cos(x)|=1,-2pi<= x<= 2pi

cos (x) + ∣ cos (x) ∣ = 1, - 2 π \leq x \leq 2 π

(30.56)/(sin(88))=(17)/(sin(x))

\frac{30.56}{sin ( 8 8 ^{\circ} )} = \frac{17}{sin ( x )}