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受欢迎的三角函数 >

sin^2(θ)=9(cos(-θ)+1)

解答

sin^{2} (θ) = 9 (cos (- θ) + 1)

解答

θ = π + 2 πn

+1

度数

θ = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

sin^{2} (θ) = 9 (cos (- θ) + 1)

使用三角恒等式改写

sin^{2} (θ) = 9 (cos (- θ) + 1)

sin^{2} (θ) = 9 (1 + cos (θ))

sin^{2} (θ) = 9 (1 + cos (θ))

两边减去

9 (1 + cos (θ))

sin^{2} (θ) - 9 (1 + cos (θ)) = 0

使用三角恒等式改写

sin^{2} (θ) - (1 + cos (θ)) \cdot 9

使用毕达哥拉斯恒等式:

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

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

= 1 - cos^{2} (θ) - (1 + cos (θ)) \cdot 9

化简

1 - cos^{2} (θ) - (1 + cos (θ)) \cdot 9 : - cos^{2} (θ) - 9 cos (θ) - 8

1 - cos^{2} (θ) - (1 + cos (θ)) \cdot 9

= 1 - cos^{2} (θ) - 9 (1 + cos (θ))

乘开

- 9 (1 + cos (θ)) : - 9 - 9 cos (θ)

- 9 (1 + cos (θ))

使用分配律:

a (b + c) = ab + a c

a = - 9, b = 1, c = cos (θ)

= - 9 \cdot 1 + (- 9) cos (θ)

使用加减运算法则

+ (- a) = - a

= - 9 \cdot 1 - 9 cos (θ)

数字相乘:

9 \cdot 1 = 9

= - 9 - 9 cos (θ)

= 1 - cos^{2} (θ) - 9 - 9 cos (θ)

化简

1 - cos^{2} (θ) - 9 - 9 cos (θ) : - cos^{2} (θ) - 9 cos (θ) - 8

1 - cos^{2} (θ) - 9 - 9 cos (θ)

对同类项分组

= - cos^{2} (θ) - 9 cos (θ) + 1 - 9

数字相加/相减:

1 - 9 = - 8

= - cos^{2} (θ) - 9 cos (θ) - 8

= - cos^{2} (θ) - 9 cos (θ) - 8

= - cos^{2} (θ) - 9 cos (θ) - 8

- 8 - cos^{2} (θ) - 9 cos (θ) = 0

用替代法求解

- 8 - cos^{2} (θ) - 9 cos (θ) = 0

令

: cos (θ) = u

- 8 - u^{2} - 9 u = 0

- 8 - u^{2} - 9 u = 0 : u = - 8, u = - 1

- 8 - u^{2} - 9 u = 0

改写成标准形式

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

- u^{2} - 9 u - 8 = 0

使用求根公式求解

- u^{2} - 9 u - 8 = 0

二次方程求根公式:

若

a = - 1, b = - 9, c = - 8

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 ( - 1 ) ( - 8 )}{2 ( - 1 )}

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 ( - 1 ) ( - 8 )}{2 ( - 1 )}

(- 9)^{2} - 4 (- 1) (- 8) = 7

(- 9)^{2} - 4 (- 1) (- 8)

使用法则

- (- a) = a

= (- 9)^{2} - 4 \cdot 1 \cdot 8

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 9)^{2} = 9^{2}

= 9^{2} - 4 \cdot 1 \cdot 8

数字相乘:

4 \cdot 1 \cdot 8 = 32

= 9^{2} - 32

9^{2} = 81

= 81 - 32

数字相减:

81 - 32 = 49

= 49

因式分解数字:

49 = 7^{2}

= 7^{2}

使用根式运算法则:

n a^{n} = a

7^{2} = 7

= 7

u_{1, 2} = \frac{- ( - 9 ) \pm 7}{2 ( - 1 )}

将解分隔开

u_{1} = \frac{- ( - 9 ) + 7}{2 ( - 1 )}, u_{2} = \frac{- ( - 9 ) - 7}{2 ( - 1 )}

u = \frac{- ( - 9 ) + 7}{2 ( - 1 )} : - 8

\frac{- ( - 9 ) + 7}{2 ( - 1 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{9 + 7}{- 2 \cdot 1}

数字相加:

9 + 7 = 16

= \frac{16}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{16}{- 2}

使用分式法则:

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

= - \frac{16}{2}

数字相除:

\frac{16}{2} = 8

= - 8

u = \frac{- ( - 9 ) - 7}{2 ( - 1 )} : - 1

\frac{- ( - 9 ) - 7}{2 ( - 1 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{9 - 7}{- 2 \cdot 1}

数字相减:

9 - 7 = 2

= \frac{2}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{2}{- 2}

使用分式法则:

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

= - \frac{2}{2}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

u = - 8, u = - 1

u = cos (θ)

代回

cos (θ) = - 8, cos (θ) = - 1

cos (θ) = - 8, cos (θ) = - 1

cos (θ) = - 8 :

无解

cos (θ) = - 8

- 1 \leq cos (x) \leq 1

无解

cos (θ) = - 1 : θ = π + 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 = π + 2 πn

θ = π + 2 πn

合并所有解

θ = π + 2 πn

作图

流行的例子

sin(30x)=202.07

sin (30 x) = 202.07

sin(x)-cos^2(x)=0

sin (x) - cos^{2} (x) = 0

solvefor x,sin(x^2)+cos(y^3)=-2

so l v e f or x, sin (x^{2}) + cos (y^{3}) = - 2

8sin(120-x)=9sin(x)

8 sin (12 0^{\circ} - x) = 9 sin (x)

cos (θ) = 0.625