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受欢迎的三角函数 >

cos(θ)=tan(θ)

解答

cos (θ) = tan (θ)

解答

θ = 0.66623 \dots + 2 πn, θ = π - 0.66623 \dots + 2 πn

+1

度数

θ = 38.17270 \dots^{\circ} + 36 0^{\circ} n, θ = 141.82729 \dots^{\circ} + 36 0^{\circ} n

求解步骤

cos (θ) = tan (θ)

两边减去

tan (θ)

cos (θ) - tan (θ) = 0

用 sin, cos 表示

cos (θ) - tan (θ)

使用基本三角恒等式:

tan (x) = \frac{sin ( x )}{cos ( x )}

= cos (θ) - \frac{sin ( θ )}{cos ( θ )}

化简

cos (θ) - \frac{sin ( θ )}{cos ( θ )} : \frac{cos ^{2} ( θ ) - sin ( θ )}{cos ( θ )}

cos (θ) - \frac{sin ( θ )}{cos ( θ )}

将项转换为分式:

cos (θ) = \frac{cos ( θ ) cos ( θ )}{cos ( θ )}

= \frac{cos ( θ ) cos ( θ )}{cos ( θ )} - \frac{sin ( θ )}{cos ( θ )}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{cos ( θ ) cos ( θ ) - sin ( θ )}{cos ( θ )}

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

cos (θ) cos (θ) - sin (θ)

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

使用指数法则:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

数字相加:

1 + 1 = 2

= cos^{2} (θ)

= cos^{2} (θ) - sin (θ)

= \frac{cos ^{2} ( θ ) - sin ( θ )}{cos ( θ )}

= \frac{cos ^{2} ( θ ) - sin ( θ )}{cos ( θ )}

\frac{cos ^{2} ( θ ) - sin ( θ )}{cos ( θ )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

cos^{2} (θ) - sin (θ) = 0

使用三角恒等式改写

cos^{2} (θ) - sin (θ)

使用毕达哥拉斯恒等式:

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

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

= 1 - sin^{2} (θ) - sin (θ)

1 - sin (θ) - sin^{2} (θ) = 0

用替代法求解

1 - sin (θ) - sin^{2} (θ) = 0

令

: sin (θ) = u

1 - u - u^{2} = 0

1 - u - u^{2} = 0 : u = - \frac{1 + 5}{2}, u = \frac{5 - 1}{2}

1 - u - u^{2} = 0

改写成标准形式

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

- u^{2} - u + 1 = 0

使用求根公式求解

- u^{2} - u + 1 = 0

二次方程求根公式:

若

a = - 1, b = - 1, c = 1

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

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

(- 1)^{2} - 4 (- 1) \cdot 1 = 5

(- 1)^{2} - 4 (- 1) \cdot 1

使用法则

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

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

若

n

是偶数

(- 1)^{2} = 1^{2}

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 4

= 1 + 4

数字相加:

1 + 4 = 5

= 5

u_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 ( - 1 )}

将解分隔开

u_{1} = \frac{- ( - 1 ) + 5}{2 ( - 1 )}, u_{2} = \frac{- ( - 1 ) - 5}{2 ( - 1 )}

u = \frac{- ( - 1 ) + 5}{2 ( - 1 )} : - \frac{1 + 5}{2}

\frac{- ( - 1 ) + 5}{2 ( - 1 )}

去除括号:

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

= \frac{1 + 5}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{1 + 5}{- 2}

使用分式法则:

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

= - \frac{1 + 5}{2}

u = \frac{- ( - 1 ) - 5}{2 ( - 1 )} : \frac{5 - 1}{2}

\frac{- ( - 1 ) - 5}{2 ( - 1 )}

去除括号:

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

= \frac{1 - 5}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{1 - 5}{- 2}

使用分式法则:

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

1 - 5 = - (5 - 1)

= \frac{5 - 1}{2}

二次方程组的解是:

u = - \frac{1 + 5}{2}, u = \frac{5 - 1}{2}

u = sin (θ)

代回

sin (θ) = - \frac{1 + 5}{2}, sin (θ) = \frac{5 - 1}{2}

sin (θ) = - \frac{1 + 5}{2}, sin (θ) = \frac{5 - 1}{2}

sin (θ) = - \frac{1 + 5}{2} :

无解

sin (θ) = - \frac{1 + 5}{2}

- 1 \leq sin (x) \leq 1

无解

sin (θ) = \frac{5 - 1}{2} : θ = arcsin (\frac{5 - 1}{2}) + 2 πn, θ = π - arcsin (\frac{5 - 1}{2}) + 2 πn

sin (θ) = \frac{5 - 1}{2}

使用反三角函数性质

sin (θ) = \frac{5 - 1}{2}

sin (θ) = \frac{5 - 1}{2}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

θ = arcsin (\frac{5 - 1}{2}) + 2 πn, θ = π - arcsin (\frac{5 - 1}{2}) + 2 πn

θ = arcsin (\frac{5 - 1}{2}) + 2 πn, θ = π - arcsin (\frac{5 - 1}{2}) + 2 πn

合并所有解

θ = arcsin (\frac{5 - 1}{2}) + 2 πn, θ = π - arcsin (\frac{5 - 1}{2}) + 2 πn

以小数形式表示解

θ = 0.66623 \dots + 2 πn, θ = π - 0.66623 \dots + 2 πn

作图

流行的例子

tan(pi/4+x)=(1-tan(x))/(1+tan(x))

tan (\frac{π}{4} + x) = \frac{1 - tan ( x )}{1 + tan ( x )}

cos (x) = \frac{45}{52}

solvefor x,cos(2x)=(m-1)/(m+1)

so l v e f or x, cos (2 x) = \frac{m - 1}{m + 1}

6 tan^{2} (x) + 8 = 10

2sin(x-pi/4)+sqrt(2)cos(x)=1

2 sin (x - \frac{π}{4}) + 2 cos (x) = 1