化简 (sin(x)-sin(y))^2+(cos(x)-cos(y))^2-2

解答

化简

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

- 2 cos (x - y)

求解步骤

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

展开

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

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

展开

(cos (x) - cos (y))^{2} : cos^{2} (x) - 2 cos (x) cos (y) + cos^{2} (y)

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

使用毕达哥拉斯恒等式:

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

= - 2 cos (x) cos (y) - 2 sin (x) sin (y) + 1 + 1 - 2

数字相加:

1 + 1 - 2 = 0

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

使用角差恒等式:

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

= - 2 cos (x - y)

因式分解出通项

2

= - 2 cos (x - y)

e x p an d (8, - 5) y (- 3, 9)

e x p an d - \frac{2 x + 2}{( x ^{2} + 2 x - 2 ) ^{2}}

e x p an d \frac{y ( 2 x ^{\frac{1}{4}} - x y ^{\frac{1}{3}} + 1 )}{2 x}

e x p an d (x - 6)^{4}

e x p an d 10 (y + 5 x) e^{y + 5 x} + e^{y + 5 x} \cdot 5 (y + 5 x)^{2}