Question

sin²(α+30º)+sin²(60º-α)-tan(45º+α)tan(45α-α)

Answer 1

问：sin²(α+30º)+sin²(60º-α)-tan(45º+α)tan(45α-α)

答：亲亲首先，我们可以使用三角恒等式将三个三角函数表示为更基本的三角函数：\sin^2(\alpha+30^\circ)=\sin^2\alpha+\sin 60^\circ\sin(2\alpha+60^\circ)sin 2 (α+30 ∘ )=sin 2 α+sin60 ∘ sin(2α+60 ∘ )\sin^2(60^\circ-\alpha)=\sin^2\alpha-\sin 60^\circ\sin(2\alpha-60^\circ)sin 2 (60 ∘ −α)=sin 2 α−sin60 ∘ sin(2α−60 ∘ )\tan(45^\circ+\alpha)\tan(45^\circ-\alpha)=\frac{\sin(45^\circ+\alpha)}{\cos(45^\circ+\alpha)}\cdot\frac{\sin(45^\circ-\alpha)}{\cos(45^\circ-\alpha)}=\frac{\sin 2\alpha}{\cos^2 45^\circ-\sin^2\alpha}=\frac{\sin 2\alpha}{1-\sin^2\alpha}tan(45 ∘ +α)tan(45 ∘ −α)= cos(45 ∘ +α)sin(45 ∘ +α)​ ⋅ cos(45 ∘ −α)sin(45 ∘ −α)​ = cos 2 45 ∘ −sin 2 αsin2α​ = 1−sin 2 αsin2α​ 将上述三个式子代入原式并化简得：\begin{aligned}&\sin^2\alpha+\sin 60^\circ\sin(2\alpha+60^\circ)+\sin^2\alpha-\sin 60^\circ\sin(2\alpha-60^\circ)-\frac{\sin 2\alpha}{1-\sin^2\alpha}\\=&2\sin^2\alpha+\sin 60^\circ\left[\sin(2\alpha+60^\circ)-\sin(2\alpha-60^\circ)\right]-\frac{\sin 2\alpha}{1-\sin^2\alpha}\\=&2\sin^2\a