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受欢迎的三角函数 >

证明 cot(θ)=(1+cos(2θ))/(sin(2θ))

解答

证明

cot (θ) = \frac{1 + cos ( 2 θ )}{sin ( 2 θ )}

解答

真

求解步骤

cot (θ) = \frac{1 + cos ( 2 θ )}{sin ( 2 θ )}

调整右侧

\frac{1 + cos ( 2 θ )}{sin ( 2 θ )}

使用三角恒等式改写

\frac{1 + cos ( 2 θ )}{sin ( 2 θ )}

使用倍角公式:

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

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

使用倍角公式:

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

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

化简

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

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

1 + 2 cos^{2} (θ) - 1 = 2 cos^{2} (θ)

1 + 2 cos^{2} (θ) - 1

对同类项分组

= 2 cos^{2} (θ) + 1 - 1

1 - 1 = 0

= 2 cos^{2} (θ)

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

数字相除:

\frac{2}{2} = 1

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

约分:

cos (θ)

= \frac{cos ( θ )}{sin ( θ )}

= \frac{cos ( θ )}{sin ( θ )}

= \frac{cos ( θ )}{sin ( θ )}

使用基本三角恒等式:

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

= cot (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 ((sin(x)))/((1-cos(x)))-((1))/((sin(x)))=cot(x)

p ro v e \frac{( sin ( x ) )}{( 1 - cos ( x ) )} - \frac{( 1 )}{( sin ( x ) )} = cot (x)

证明 sin(0)sec(0)=tan(0)

p ro v e sin (0) sec (0) = tan (0)

证明 cos(2a)=(cot(a)-tan(a))/(csc(a)sec(a))

p ro v e cos (2 a) = \frac{cot ( a ) - tan ( a )}{csc ( a ) sec ( a )}

证明 cos(a)cos(b)= 1/2 (cos(a-b)+cos(a+b))

p ro v e cos (a) cos (b) = \frac{1}{2} (cos (a - b) + cos (a + b))

证明 sin(a-b)=sin(a)cos(b)-cos(a)sin(b)

p ro v e sin (a - b) = sin (a) cos (b) - cos (a) sin (b)