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受欢迎的三角函数 >

证明 (1+sin(x))^2+cos^2(x)=2+2sin(x)

解答

证明

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

解答

真

求解步骤

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

调整左侧

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

化简

(1 + sin (x))^{2} + 1 - sin^{2} (x) : 2 sin (x) + 2

(1 + sin (x))^{2} + 1 - sin^{2} (x)

(1 + sin (x))^{2} : 1 + 2 sin (x) + sin^{2} (x)

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = 1, b = sin (x)

= 1^{2} + 2 \cdot 1 \cdot sin (x) + sin^{2} (x)

化简

1^{2} + 2 \cdot 1 \cdot sin (x) + sin^{2} (x) : 1 + 2 sin (x) + sin^{2} (x)

1^{2} + 2 \cdot 1 \cdot sin (x) + sin^{2} (x)

使用法则

1^{a} = 1

1^{2} = 1

= 1 + 2 \cdot 1 \cdot sin (x) + sin^{2} (x)

数字相乘:

2 \cdot 1 = 2

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

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

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

化简

1 + 2 sin (x) + sin^{2} (x) + 1 - sin^{2} (x) : 2 sin (x) + 2

1 + 2 sin (x) + sin^{2} (x) + 1 - sin^{2} (x)

对同类项分组

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

同类项相加:

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

= 2 sin (x) + 1 + 1

数字相加:

1 + 1 = 2

= 2 sin (x) + 2

= 2 sin (x) + 2

= 2 sin (x) + 2

= 2 sin (x) + 2

= 2 + 2 sin (x)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

流行的例子

证明 cot(60)=(cos(60))/(sin(60))

p ro v e cot (6 0^{\circ}) = \frac{cos ( 6 0 ^{\circ} )}{sin ( 6 0 ^{\circ} )}

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p ro v e tan (- x) tan (\frac{π}{2} - x) = - 1

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p ro v e tan (π - θ) = - tan (x)

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p ro v e cot (θ) (sin (θ) + tan (θ)) = cos (θ) + 1

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p ro v e (2 - sin^{2} (x)) csc^{2} (x) = cot^{2} (x)