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受欢迎的三角函数 >

solvefor θ,r^2cos^2(θ)+r^2sin^2(θ)<= 1

解答

求解

θ, r^{2} cos^{2} (θ) + r^{2} sin^{2} (θ) \leq 1

解答

- 1 \leq θ \leq 1

求解步骤

r^{2} cos^{2} (θ) + r^{2} sin^{2} (θ) \leq 1

利用以下特性:

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

因此

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

r^{2} (1 - sin^{2} (θ)) + r^{2} sin^{2} (θ) \leq 1

化简

r^{2} (1 - sin^{2} (θ)) + r^{2} sin^{2} (θ) : r^{2}

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

乘开

r^{2} (1 - sin^{2} (θ)) : r^{2} - r^{2} sin^{2} (θ)

r^{2} (1 - sin^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = r^{2}, b = 1, c = sin^{2} (θ)

= r^{2} \cdot 1 - r^{2} sin^{2} (θ)

= 1 \cdot r^{2} - r^{2} sin^{2} (θ)

乘以:

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

= r^{2} - r^{2} sin^{2} (θ)

= r^{2} - r^{2} sin^{2} (θ) + r^{2} sin^{2} (θ)

同类项相加:

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

= r^{2}

r^{2} \leq 1

改写为标准形式

r^{2} \leq 1

两边减去

1

r^{2} - 1 \leq 1 - 1

化简

r^{2} - 1 \leq 0

r^{2} - 1 \leq 0

分解

r^{2} - 1 : (r + 1) (r - 1)

r^{2} - 1

将

1

改写为

1^{2}

= r^{2} - 1^{2}

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

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

= (r + 1) (r - 1)

(r + 1) (r - 1) \leq 0

确定区间

确定

(r + 1) (r - 1)

符号

确定

r + 1

符号

r + 1 = 0 : r = - 1

r + 1 = 0

将

1

到右边

r + 1 = 0

两边减去

1

r + 1 - 1 = 0 - 1

化简

r = - 1

r = - 1

r + 1 < 0 : r < - 1

r + 1 < 0

将

1

到右边

r + 1 < 0

两边减去

1

r + 1 - 1 < 0 - 1

化简

r < - 1

r < - 1

r + 1 > 0 : r > - 1

r + 1 > 0

将

1

到右边

r + 1 > 0

两边减去

1

r + 1 - 1 > 0 - 1

化简

r > - 1

r > - 1

确定

r - 1

符号

r - 1 = 0 : r = 1

r - 1 = 0

将

1

到右边

r - 1 = 0

两边加上

1

r - 1 + 1 = 0 + 1

化简

r = 1

r = 1

r - 1 < 0 : r < 1

r - 1 < 0

将

1

到右边

r - 1 < 0

两边加上

1

r - 1 + 1 < 0 + 1

化简

r < 1

r < 1

r - 1 > 0 : r > 1

r - 1 > 0

将

1

到右边

r - 1 > 0

两边加上

1

r - 1 + 1 > 0 + 1

化简

r > 1

r > 1

总结如下表：

r + 1 r - 1 (r + 1) (r - 1) r < - 1 - - + r = - 1 0 - 0 - 1 < r < 1 + - - r = 1 + 00 r > 1 + + +

确定满足所需条件的区间：

\leq 0

r = - 1 or - 1 < r < 1 or r = 1

合并重叠的区间

- 1 \leq r < 1 or r = 1

两个区间的并集是指存在于任一区间的数的集合

r = - 1

or

- 1 < r < 1

- 1 \leq r < 1

两个区间的并集是指存在于任一区间的数的集合

- 1 \leq r < 1

or

r = 1

- 1 \leq r \leq 1

- 1 \leq r \leq 1

- 1 \leq θ \leq 1

流行的例子

1-2cos(2x)>sin^2(x)

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

cos^{2} (x) \geq \frac{1}{2}

cos^2(x)+(sqrt(2))/2 cos(x)>0

cos^{2} (x) + \frac{2}{2} cos (x) > 0

sin (x) \geq 0.5

cos(3x)<= (sqrt(3))/2

cos (3 x) \leq \frac{3}{2}