zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

证明 7cos^2(x)+5sin^2(x)=6+cos(2x)

解答

证明

7 cos^{2} (x) + 5 sin^{2} (x) = 6 + cos (2 x)

解答

真

求解步骤

7 cos^{2} (x) + 5 sin^{2} (x) = 6 + cos (2 x)

调整左侧

7 cos^{2} (x) + 5 sin^{2} (x)

使用三角恒等式改写

7 cos^{2} (x) + 5 sin^{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)

= 7 (1 - sin^{2} (x)) + 5 sin^{2} (x)

化简

7 (1 - sin^{2} (x)) + 5 sin^{2} (x) : 7 - 2 sin^{2} (x)

7 (1 - sin^{2} (x)) + 5 sin^{2} (x)

乘开

7 (1 - sin^{2} (x)) : 7 - 7 sin^{2} (x)

7 (1 - sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 7, b = 1, c = sin^{2} (x)

= 7 \cdot 1 - 7 sin^{2} (x)

数字相乘:

7 \cdot 1 = 7

= 7 - 7 sin^{2} (x)

= 7 - 7 sin^{2} (x) + 5 sin^{2} (x)

同类项相加:

- 7 sin^{2} (x) + 5 sin^{2} (x) = - 2 sin^{2} (x)

= 7 - 2 sin^{2} (x)

= 7 - 2 sin^{2} (x)

= 7 - 2 sin^{2} (x)

调整右侧

6 + cos (2 x)

使用三角恒等式改写

6 + cos (2 x)

使用倍角公式:

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

= 6 + 1 - 2 sin^{2} (x)

化简

= - 2 sin^{2} (x) + 7

= - 2 sin^{2} (x) + 7

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 2/(1-cos(2x))=csc^2(x)

p ro v e \frac{2}{1 - cos ( 2 x )} = csc^{2} (x)

证明 cos(10x)-cos(6x)=-2sin(8x)sin(2x)

p ro v e cos (10 x) - cos (6 x) = - 2 sin (8 x) sin (2 x)

证明 (1+cos(2x))/(cos(x))=2cos(x)

p ro v e \frac{1 + cos ( 2 x )}{cos ( x )} = 2 cos (x)

证明 sec(y)-cos(y)=tan(y)sin(y)

p ro v e sec (y) - cos (y) = tan (y) sin (y)

证明 sin(5x)-sin(3x)=2cos(4x)sin(x)

p ro v e sin (5 x) - sin (3 x) = 2 cos (4 x) sin (x)