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受欢迎的三角函数 >

2cos^2(t)+2cos(2t)=0

解答

2 cos^{2} (t) + 2 cos (2 t) = 0

解答

t = - 0.95531 \dots + πn, t = 0.95531 \dots + πn

+1

度数

t = - 54.73561 \dots^{\circ} + 18 0^{\circ} n, t = 54.73561 \dots^{\circ} + 18 0^{\circ} n

求解步骤

2 cos^{2} (t) + 2 cos (2 t) = 0

使用三角恒等式改写

2 cos (2 t) + 2 cos^{2} (t)

使用倍角公式:

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

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

化简

2 (cos^{2} (t) - sin^{2} (t)) + 2 cos^{2} (t) : 4 cos^{2} (t) - 2 sin^{2} (t)

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

乘开

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

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

使用分配律:

a (b - c) = ab - a c

a = 2, b = cos^{2} (t), c = sin^{2} (t)

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

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

同类项相加:

2 cos^{2} (t) + 2 cos^{2} (t) = 4 cos^{2} (t)

= 4 cos^{2} (t) - 2 sin^{2} (t)

= 4 cos^{2} (t) - 2 sin^{2} (t)

- 2 sin^{2} (t) + 4 cos^{2} (t) = 0

分解

- 2 sin^{2} (t) + 4 cos^{2} (t) : 2 (2 cos (t) + sin (t)) (2 cos (t) - sin (t))

- 2 sin^{2} (t) + 4 cos^{2} (t)

将

4

改写为

2 \cdot 2

= - 2 sin^{2} (t) + 2 \cdot 2 cos^{2} (t)

因式分解出通项

2

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

分解

2 cos^{2} (t) - sin^{2} (t) : (2 cos (t) + sin (t)) (2 cos (t) - sin (t))

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

将

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

改写为

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

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

使用根式运算法则:

a = (a)^{2}

2 = (2)^{2}

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

使用指数法则:

a^{m} b^{m} = (ab)^{m}

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

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

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

使用平方差公式:

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

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

= (2 cos (t) + sin (t)) (2 cos (t) - sin (t))

= 2 (2 cos (t) + sin (t)) (2 cos (t) - sin (t))

2 (2 cos (t) + sin (t)) (2 cos (t) - sin (t)) = 0

分别求解每个部分

2 cos (t) + sin (t) = 0 or 2 cos (t) - sin (t) = 0

2 cos (t) + sin (t) = 0 : t = arctan (- 2) + πn

2 cos (t) + sin (t) = 0

使用三角恒等式改写

2 cos (t) + sin (t) = 0

在两边除以

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

\frac{2 cos ( t ) + sin ( t )}{cos ( t )} = \frac{0}{cos ( t )}

化简

2 + \frac{sin ( t )}{cos ( t )} = 0

使用基本三角恒等式:

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

2 + tan (t) = 0

2 + tan (t) = 0

将

2

到右边

2 + tan (t) = 0

两边减去

2

2 + tan (t) - 2 = 0 - 2

化简

tan (t) = - 2

tan (t) = - 2

使用反三角函数性质

tan (t) = - 2

tan (t) = - 2

的通解

tan (x) = - a \Rightarrow x = arctan (- a) + πn

t = arctan (- 2) + πn

t = arctan (- 2) + πn

2 cos (t) - sin (t) = 0 : t = arctan (2) + πn

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

使用三角恒等式改写

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

在两边除以

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

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

化简

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

使用基本三角恒等式:

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

2 - tan (t) = 0

2 - tan (t) = 0

将

2

到右边

2 - tan (t) = 0

两边减去

2

2 - tan (t) - 2 = 0 - 2

化简

- tan (t) = - 2

- tan (t) = - 2

两边除以

- 1

- tan (t) = - 2

两边除以

- 1

\frac{- tan ( t )}{- 1} = \frac{- 2}{- 1}

化简

tan (t) = 2

tan (t) = 2

使用反三角函数性质

tan (t) = 2

tan (t) = 2

的通解

tan (x) = a \Rightarrow x = arctan (a) + πn

t = arctan (2) + πn

t = arctan (2) + πn

合并所有解

t = arctan (- 2) + πn, t = arctan (2) + πn

以小数形式表示解

t = - 0.95531 \dots + πn, t = 0.95531 \dots + πn

作图

流行的例子

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arcsin (2 x) + arcsin (x) = \frac{2 π}{3}

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2 csc (x) + 2 = 0

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tan (2 c) = - tan (x)