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受欢迎的三角函数 >

sec^2(θ)=1-tan(θ)

解答

sec^{2} (θ) = 1 - tan (θ)

解答

θ = \frac{3 π}{4} + πn

+1

度数

θ = 13 5^{\circ} + 18 0^{\circ} n

求解步骤

sec^{2} (θ) = 1 - tan (θ)

两边减去

1 - tan (θ)

sec^{2} (θ) - 1 + tan (θ) = 0

使用三角恒等式改写

- 1 + sec^{2} (θ) + tan (θ)

使用毕达哥拉斯恒等式:

sec^{2} (x) = tan^{2} (x) + 1

sec^{2} (x) - 1 = tan^{2} (x)

= tan (θ) + tan^{2} (θ)

使用基本三角恒等式:

tan (x) = \frac{1}{cot ( x )}

= \frac{1}{cot ( θ )} + (\frac{1}{cot ( θ )})^{2}

(\frac{1}{cot ( θ )})^{2} = \frac{1}{cot ^{2} ( θ )}

(\frac{1}{cot ( θ )})^{2}

使用指数法则:

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

= \frac{1 ^{2}}{cot ^{2} ( θ )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{cot ^{2} ( θ )}

= \frac{1}{cot ( θ )} + \frac{1}{cot ^{2} ( θ )}

\frac{1}{cot ^{2} ( θ )} + \frac{1}{cot ( θ )} = 0

用替代法求解

\frac{1}{cot ^{2} ( θ )} + \frac{1}{cot ( θ )} = 0

令

: cot (θ) = u

\frac{1}{u ^{2}} + \frac{1}{u} = 0

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

\frac{1}{u ^{2}} + \frac{1}{u} = 0

乘以最小公倍数

\frac{1}{u ^{2}} + \frac{1}{u} = 0

找到

u^{2}, u

的最小公倍数:

u^{2}

u^{2}, u

最小公倍数 (LCM)

计算出由出现在

u^{2}

或

u

中的因子组成的表达式

= u^{2}

乘以最小公倍数=

u^{2}

\frac{1}{u ^{2}} u^{2} + \frac{1}{u} u^{2} = 0 \cdot u^{2}

化简

\frac{1}{u ^{2}} u^{2} + \frac{1}{u} u^{2} = 0 \cdot u^{2}

化简

\frac{1}{u ^{2}} u^{2} : 1

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

分式相乘:

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

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

约分:

u^{2}

= 1

化简

\frac{1}{u} u^{2} : u

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

分式相乘:

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

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

乘以:

1 \cdot u^{2} = u^{2}

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

约分:

u

= u

化简

0 \cdot u^{2} : 0

0 \cdot u^{2}

使用法则

0 \cdot a = 0

= 0

1 + u = 0

1 + u = 0

1 + u = 0

将

1

到右边

1 + u = 0

两边减去

1

1 + u - 1 = 0 - 1

化简

u = - 1

u = - 1

验证解

找到无定义的点（奇点）:

u = 0

取

\frac{1}{u ^{2}} + \frac{1}{u}

的分母，令其等于零

解

u^{2} = 0 : u = 0

u^{2} = 0

使用法则

x^{n} = 0 \Rightarrow x = 0

u = 0

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

u = - 1

u = cot (θ)

代回

cot (θ) = - 1

cot (θ) = - 1

cot (θ) = - 1 : θ = \frac{3 π}{4} + πn

cot (θ) = - 1

cot (θ) = - 1

的通解

cot (x)

周期表（周期为

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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

合并所有解

θ = \frac{3 π}{4} + πn

作图

流行的例子

cos^2(x)=-0.267

cos^{2} (x) = - 0.267

sin(x-4)=cos(9x+4)

sin (x - 4) = cos (9 x + 4)

solvefor x,0=-2sin(x)-4cos(2x)

so l v e f or x, 0 = - 2 sin (x) - 4 cos (2 x)

cos^2(θ)=2cos(θ)

cos^{2} (θ) = 2 cos (θ)

2sin(x)-sin(2x)=sin(2x)

2 sin (x) - sin (2 x) = sin (2 x)