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受欢迎的三角函数 >

10tan(θ)-4.9((10)/(13cos(θ)))^2=0

解答

10 tan (θ) - 4.9 (\frac{10}{13 cos ( θ )})^{2} = 0

解答

θ = \frac{0.61858 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.61858 \dots}{2} + πn

+1

度数

θ = 17.72110 \dots^{\circ} + 18 0^{\circ} n, θ = 72.27889 \dots^{\circ} + 18 0^{\circ} n

求解步骤

10 tan (θ) - 4.9 (\frac{10}{13 cos ( θ )})^{2} = 0

用 sin, cos 表示

- (\frac{10}{13 cos ( θ )})^{2} \cdot 4.9 + 10 tan (θ)

使用基本三角恒等式:

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

= - (\frac{10}{13 cos ( θ )})^{2} \cdot 4.9 + 10 \cdot \frac{sin ( θ )}{cos ( θ )}

化简

- (\frac{10}{13 cos ( θ )})^{2} \cdot 4.9 + 10 \cdot \frac{sin ( θ )}{cos ( θ )} : \frac{- 2.89940 \dots + 10 sin ( θ ) cos ( θ )}{cos ^{2} ( θ )}

- (\frac{10}{13 cos ( θ )})^{2} \cdot 4.9 + 10 \cdot \frac{sin ( θ )}{cos ( θ )}

(\frac{10}{13 cos ( θ )})^{2} \cdot 4.9 = \frac{2.89940 \dots}{cos ^{2} ( θ )}

(\frac{10}{13 cos ( θ )})^{2} \cdot 4.9

(\frac{10}{13 cos ( θ )})^{2} = \frac{1 0 ^{2}}{1 3 ^{2} cos ^{2} ( θ )}

(\frac{10}{13 cos ( θ )})^{2}

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 0 ^{2}}{( 13 cos ( θ ) ) ^{2}}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

(13 cos (θ))^{2} = 1 3^{2} cos^{2} (θ)

= \frac{1 0 ^{2}}{1 3 ^{2} cos ^{2} ( θ )}

= 4.9 \cdot \frac{1 0 ^{2}}{1 3 ^{2} cos ^{2} ( θ )}

分式相乘:

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

= \frac{1 0 ^{2} \cdot 4.9}{1 3 ^{2} cos ^{2} ( θ )}

1 0^{2} \cdot 4.9 = 490

1 0^{2} \cdot 4.9

1 0^{2} = 100

= 100 \cdot 4.9

数字相乘:

100 \cdot 4.9 = 490

= 490

= \frac{490}{1 3 ^{2} cos ^{2} ( θ )}

1 3^{2} = 169

= \frac{490}{169 cos ^{2} ( θ )}

转换为小数形式

\frac{490}{169} = 2.89940 \dots

= \frac{2.89940 \dots}{cos ^{2} ( θ )}

10 \cdot \frac{sin ( θ )}{cos ( θ )} = \frac{10 sin ( θ )}{cos ( θ )}

10 \cdot \frac{sin ( θ )}{cos ( θ )}

分式相乘:

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

= \frac{sin ( θ ) \cdot 10}{cos ( θ )}

= - \frac{2.89940 \dots}{cos ^{2} ( θ )} + \frac{10 sin ( θ )}{cos ( θ )}

cos^{2} (θ), cos (θ)

的最小公倍数:

cos^{2} (θ)

cos^{2} (θ), cos (θ)

最小公倍数 (LCM)

计算出由出现在

cos^{2} (θ)

或

cos (θ)

中的因子组成的表达式

= cos^{2} (θ)

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

cos^{2} (θ)

对于

\frac{sin ( θ ) \cdot 10}{cos ( θ )} :

将分母和分子乘以

cos (θ)

\frac{sin ( θ ) \cdot 10}{cos ( θ )} = \frac{sin ( θ ) \cdot 10 cos ( θ )}{cos ( θ ) cos ( θ )} = \frac{sin ( θ ) \cdot 10 cos ( θ )}{cos ^{2} ( θ )}

= - \frac{2.89940 \dots}{cos ^{2} ( θ )} + \frac{sin ( θ ) \cdot 10 cos ( θ )}{cos ^{2} ( 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{- 2.89940 \dots + sin ( θ ) \cdot 10 cos ( θ )}{cos ^{2} ( θ )}

= \frac{- 2.89940 \dots + 10 sin ( θ ) cos ( θ )}{cos ^{2} ( θ )}

\frac{- 2.89940 \dots + 10 cos ( θ ) sin ( θ )}{cos ^{2} ( θ )} = 0

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

- 2.89940 \dots + 10 cos (θ) sin (θ) = 0

使用三角恒等式改写

- 2.89940 \dots + 10 cos (θ) sin (θ)

使用倍角公式:

2 sin (x) cos (x) = sin (2 x)

sin (x) cos (x) = \frac{sin ( 2 x )}{2}

= - 2.89940 \dots + 10 \cdot \frac{sin ( 2 θ )}{2}

- 2.89940 \dots + 10 \cdot \frac{sin ( 2 θ )}{2} = 0

10 \cdot \frac{sin ( 2 θ )}{2} = 5 sin (2 θ)

10 \cdot \frac{sin ( 2 θ )}{2}

分式相乘:

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

= \frac{sin ( 2 θ ) \cdot 10}{2}

数字相除:

\frac{10}{2} = 5

= 5 sin (2 θ)

- 2.89940 \dots + 5 sin (2 θ) = 0

将

2.89940 \dots

到右边

- 2.89940 \dots + 5 sin (2 θ) = 0

两边加上

2.89940 \dots

- 2.89940 \dots + 5 sin (2 θ) + 2.89940 \dots = 0 + 2.89940 \dots

化简

5 sin (2 θ) = 2.89940 \dots

5 sin (2 θ) = 2.89940 \dots

两边除以

5

5 sin (2 θ) = 2.89940 \dots

两边除以

5

\frac{5 sin ( 2 θ )}{5} = \frac{2.89940 \dots}{5}

化简

sin (2 θ) = 0.57988 \dots

sin (2 θ) = 0.57988 \dots

使用反三角函数性质

sin (2 θ) = 0.57988 \dots

sin (2 θ) = 0.57988 \dots

的通解

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

2 θ = arcsin (0.57988 \dots) + 2 πn, 2 θ = π - arcsin (0.57988 \dots) + 2 πn

2 θ = arcsin (0.57988 \dots) + 2 πn, 2 θ = π - arcsin (0.57988 \dots) + 2 πn

解

2 θ = arcsin (0.57988 \dots) + 2 πn : θ = \frac{arcsin ( 0.57988 \dots )}{2} + πn

2 θ = arcsin (0.57988 \dots) + 2 πn

两边除以

2

2 θ = arcsin (0.57988 \dots) + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{arcsin ( 0.57988 \dots )}{2} + \frac{2 πn}{2}

化简

θ = \frac{arcsin ( 0.57988 \dots )}{2} + πn

θ = \frac{arcsin ( 0.57988 \dots )}{2} + πn

解

2 θ = π - arcsin (0.57988 \dots) + 2 πn : θ = \frac{π}{2} - \frac{arcsin ( 0.57988 \dots )}{2} + πn

2 θ = π - arcsin (0.57988 \dots) + 2 πn

两边除以

2

2 θ = π - arcsin (0.57988 \dots) + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{π}{2} - \frac{arcsin ( 0.57988 \dots )}{2} + \frac{2 πn}{2}

化简

θ = \frac{π}{2} - \frac{arcsin ( 0.57988 \dots )}{2} + πn

θ = \frac{π}{2} - \frac{arcsin ( 0.57988 \dots )}{2} + πn

θ = \frac{arcsin ( 0.57988 \dots )}{2} + πn, θ = \frac{π}{2} - \frac{arcsin ( 0.57988 \dots )}{2} + πn

以小数形式表示解

θ = \frac{0.61858 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.61858 \dots}{2} + πn

作图

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