证明 cos^2(6θ)-sin^2(6θ)=cos(12θ)

解答

证明

cos^{2} (6 θ) - sin^{2} (6 θ) = cos (12 θ)

真

求解步骤

cos^{2} (6 θ) - sin^{2} (6 θ) = cos (12 θ)

调整左侧

cos^{2} (6 θ) - sin^{2} (6 θ)

使用三角恒等式改写

cos^{2} (6 θ) - sin^{2} (6 θ)

使用倍角公式:

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

= cos (2 \cdot 6 θ)

化简

= cos (12 θ)

= cos (12 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 ( β )}{tan ( β )} = csc (β) - sin (β)

p ro v e cos (3 t) = 4 cos^{3} (t) - 3 cos (t)

p ro v e sec (2 x) = \frac{sec ^{2} ( x ) + sec ^{4} ( x )}{2 + sec ^{2} ( x ) - sec ^{4} ( x )}

p ro v e \frac{1}{csc ( x )} = \frac{1}{sin ( x )}

p ro v e 1 + tan^{2} (A) = sec^{3} (A) cos (A)