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受欢迎的三角函数 >

证明 csc(θ)(1-cos^2(θ))=sin(θ)

解答

证明

csc (θ) (1 - cos^{2} (θ)) = sin (θ)

解答

真

求解步骤

csc (θ) (1 - cos^{2} (θ)) = sin (θ)

调整左侧

csc (θ) (1 - cos^{2} (θ))

用 sin, cos 表示

(1 - cos^{2} (θ)) csc (θ)

使用基本三角恒等式:

csc (x) = \frac{1}{sin ( x )}

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

化简

(1 - cos^{2} (θ)) \frac{1}{sin ( θ )} : \frac{1 - cos ^{2} ( θ )}{sin ( θ )}

(1 - cos^{2} (θ)) \frac{1}{sin ( θ )}

分式相乘:

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

= \frac{1 \cdot ( 1 - cos ^{2} ( θ ) )}{sin ( θ )}

1 \cdot (1 - cos^{2} (θ)) = 1 - cos^{2} (θ)

1 \cdot (1 - cos^{2} (θ))

乘以:

1 \cdot (1 - cos^{2} (θ)) = (1 - cos^{2} (θ))

= (1 - cos^{2} (θ))

去除括号:

(a) = a

= 1 - cos^{2} (θ)

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

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

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

使用三角恒等式改写

\frac{1 - cos ^{2} ( θ )}{sin ( θ )}

使用毕达哥拉斯恒等式:

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

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

= \frac{sin ^{2} ( θ )}{sin ( θ )}

约分:

sin (θ)

= sin (θ)

= sin (θ)

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 ((csc^2(x)))/(cot(x))=csc(x)sec(x)

p ro v e \frac{( csc ^{2} ( x ) )}{cot ( x )} = csc (x) sec (x)

证明 csc^2(x)(tan^2(x))-1=tan^2(x)

p ro v e csc^{2} (x) (tan^{2} (x)) - 1 = tan^{2} (x)

证明 sin(x)cos(x)+sin(3x)sec(x)=tan(x)

p ro v e sin (x) cos (x) + sin (3 x) sec (x) = tan (x)

证明 (1+sec(x))(1-cos(x))=sin(x)tan(x)

p ro v e (1 + sec (x)) (1 - cos (x)) = sin (x) tan (x)

证明 sec(x)*sin(x)=tan(x)

p ro v e sec (x) \cdot sin (x) = tan (x)