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受欢迎的三角函数 >

证明 (2cos^2(x))/(sin(2x))= 1/(tan(x))

解答

证明

\frac{2 cos ^{2} ( x )}{sin ( 2 x )} = \frac{1}{tan ( x )}

解答

真

求解步骤

\frac{2 cos ^{2} ( x )}{sin ( 2 x )} = \frac{1}{tan ( x )}

调整左侧

\frac{2 cos ^{2} ( x )}{sin ( 2 x )}

使用三角恒等式改写

\frac{2 cos ^{2} ( x )}{sin ( 2 x )}

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

= \frac{2 cos ^{2} ( x )}{2 sin ( x ) cos ( x )}

化简

\frac{2 cos ^{2} ( x )}{2 sin ( x ) cos ( x )} : \frac{cos ( x )}{sin ( x )}

\frac{2 cos ^{2} ( x )}{2 sin ( x ) cos ( x )}

数字相除:

\frac{2}{2} = 1

= \frac{cos ^{2} ( x )}{sin ( x ) cos ( x )}

约分:

cos (x)

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

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

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

调整右侧

\frac{1}{tan ( x )}

用 sin, cos 表示

\frac{1}{tan ( x )}

使用基本三角恒等式:

tan (x) = \frac{sin ( x )}{cos ( x )}

= \frac{1}{\frac{s i n ( x )}{c o s ( x )}}

化简

\frac{1}{\frac{s i n ( x )}{c o s ( x )}} : \frac{cos ( x )}{sin ( x )}

\frac{1}{\frac{s i n ( x )}{c o s ( x )}}

使用分式法则:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

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

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

p ro v e cos (- 2 θ) = cos (2 θ)

证明 1-cos^2(x)=tan(x)

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

证明 cot(2x)=(1/2)(cot(x)-tan(x))

p ro v e cot (2 x) = (\frac{1}{2}) (cot (x) - tan (x))

证明 sec^2(x)cot(x)+cot(x)=tan(x)

p ro v e sec^{2} (x) cot (x) + cot (x) = tan (x)

证明 1-(cos(x))^2=1+cos^2(x)

p ro v e 1 - (cos (x))^{2} = 1 + cos^{2} (x)