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受欢迎的三角函数 >

cos^2(x)= 1/2 sin(2x)

解答

cos^{2} (x) = \frac{1}{2} sin (2 x)

解答

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{π}{4} + πn

+1

度数

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

求解步骤

cos^{2} (x) = \frac{1}{2} sin (2 x)

两边减去

\frac{1}{2} sin (2 x)

cos^{2} (x) - \frac{1}{2} sin (2 x) = 0

化简

cos^{2} (x) - \frac{1}{2} sin (2 x) : \frac{2 cos ^{2} ( x ) - sin ( 2 x )}{2}

cos^{2} (x) - \frac{1}{2} sin (2 x)

\frac{1}{2} sin (2 x) = \frac{sin ( 2 x )}{2}

\frac{1}{2} sin (2 x)

分式相乘:

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

= \frac{1 \cdot sin ( 2 x )}{2}

乘以:

1 \cdot sin (2 x) = sin (2 x)

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

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

将项转换为分式:

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

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

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 ^{2} ( x ) \cdot 2 - sin ( 2 x )}{2}

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

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

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

使用三角恒等式改写

- sin (2 x) + 2 cos^{2} (x)

使用倍角公式:

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

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

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

分解

2 cos^{2} (x) - 2 cos (x) sin (x) : 2 cos (x) (cos (x) - sin (x))

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

使用指数法则:

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

cos^{2} (x) = cos (x) cos (x)

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

因式分解出通项

2 cos (x)

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

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

分别求解每个部分

cos (x) = 0 or cos (x) - sin (x) = 0

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

cos (x) = 0

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}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) - sin (x) = 0 : x = \frac{π}{4} + πn

cos (x) - sin (x) = 0

使用三角恒等式改写

cos (x) - sin (x) = 0

在两边除以

cos (x), cos (x) \neq = 0

\frac{cos ( x ) - sin ( x )}{cos ( x )} = \frac{0}{cos ( x )}

化简

1 - \frac{sin ( x )}{cos ( x )} = 0

使用基本三角恒等式:

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

1 - tan (x) = 0

1 - tan (x) = 0

将

1

到右边

1 - tan (x) = 0

两边减去

1

1 - tan (x) - 1 = 0 - 1

化简

- tan (x) = - 1

- tan (x) = - 1

两边除以

- 1

- tan (x) = - 1

两边除以

- 1

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

化简

tan (x) = 1

tan (x) = 1

tan (x) = 1

的通解

tan (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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

合并所有解

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{π}{4} + πn

作图

流行的例子

3 cot (θ) + 5 = 0

so l v e f or x, h = g

9cos(2x)=10sin(x)+1

9 cos (2 x) = 10 sin (x) + 1

10cos(B)+6=4cos(B)+4

10 cos (B) + 6 = 4 cos (B) + 4

solvefor y,x=x(1+tan(y))

so l v e f or y, x = x (1 + tan (y))