证明 cos^2(θ)=(1+cos(2θ))/2

解答

证明

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

真

求解步骤

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

调整右侧

\frac{1 + cos ( 2 θ )}{2}

使用三角恒等式改写

\frac{1 + cos ( 2 θ )}{2}

使用倍角公式:

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

= \frac{1 + 2 cos ^{2} ( θ ) - 1}{2}

化简

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

\frac{1 + 2 cos ^{2} ( θ ) - 1}{2}

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}

数字相除:

\frac{2}{2} = 1

= cos^{2} (θ)

= cos^{2} (θ)

= cos^{2} (θ)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

p ro v e \frac{cos ^{2} ( α )}{1 - sin ( α )} = 1 + sin (α)

p ro v e 2 sin^{2} (θ) - 1 = 1 - 2 cos^{2} (θ)

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

p ro v e 2 cos^{2} (A) - 1 = 1 - 2 sin^{2} (A)

p ro v e sec^{4} (t) - tan^{4} (t) = 1 + 2 tan^{2} (t)