证明 1-sin(y)=(cos^2(y))/(1+sin(y))

解答

证明

1 - sin (y) = \frac{cos ^{2} ( y )}{1 + sin ( y )}

真

求解步骤

1 - sin (y) = \frac{cos ^{2} ( y )}{1 + sin ( y )}

调整左侧

1 - sin (y)

乘以

\frac{1 + sin ( y )}{1 + sin ( y )}

= \frac{( 1 - sin ( y ) ) ( 1 + sin ( y ) )}{1 + sin ( y )}

使用平方差公式:

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

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

= \frac{1 - sin ^{2} ( y )}{1 + sin ( y )}

使用毕达哥拉斯恒等式:

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

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

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

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

p ro v e 2 tan (x) = \frac{2 sin ( x )}{cos ( x )}

p ro v e \frac{1}{1 - sin ( B )} = sec^{2} (B) + sec (B) tan (B)

p ro v e tan (6 0^{\circ}) = \frac{2 tan ( 3 0 ^{\circ} )}{1 - tan ^{2} ( 3 0 ^{\circ} )}

p ro v e sin^{2} (u) (cot^{2} (u) + 1) = 1

p ro v e \frac{cos ( - x )}{sin ( - x )} = - cot (x)