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受欢迎的三角函数 >

证明 sec(x)cos(x)+tan^2(x)=sec^2(x)

解答

证明

sec (x) cos (x) + tan^{2} (x) = sec^{2} (x)

解答

真

求解步骤

sec (x) cos (x) + tan^{2} (x) = sec^{2} (x)

调整左侧

sec (x) cos (x) + tan^{2} (x)

使用三角恒等式改写

sec (x) cos (x) + tan^{2} (x)

sec (x) cos (x) = 1

sec (x) cos (x)

用 sin, cos 表示

sec (x) cos (x)

使用基本三角恒等式:

sec (x) = \frac{1}{cos ( x )}

= \frac{1}{cos ( x )} cos (x)

\frac{1}{cos ( x )} cos (x) = 1

\frac{1}{cos ( x )} cos (x)

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 cos ( x )}{cos ( x )}

约分:

cos (x)

= 1

= 1

= 1 + tan^{2} (x)

使用毕达哥拉斯恒等式:

tan^{2} (x) + 1 = sec^{2} (x)

= sec^{2} (x)

= sec^{2} (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^2(θ)+1= 1/(sin^2(θ))

p ro v e cot^{2} (θ) + 1 = \frac{1}{sin ^{2} ( θ )}

证明 3cos(2x+pi/2)-2=6cos^2(x+pi/4)

p ro v e 3 cos (2 x + \frac{π}{2}) - 2 = 6 cos^{2} (x + \frac{π}{4})

证明 cos^4(b)-sin^4(b)=cos(2b)

p ro v e cos^{4} (b) - sin^{4} (b) = cos (2 b)

证明 sin^2(x)sec^2(x)=sec^2(x)-1

p ro v e sin^{2} (x) sec^{2} (x) = sec^{2} (x) - 1

证明 (6cos^2(x)-6sin^2(x))=6-12sin^2(x)

p ro v e (6 cos^{2} (x) - 6 sin^{2} (x)) = 6 - 12 sin^{2} (x)