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在三角中,三角函数连乘积的证明、化简是一个难点。例如,“求证sin20°·sin40°·sin60°·sin80°=3/(16)”,一般需几次应用积化和差公式才能证得。仔细观察求证式,左端除了60°这个特殊角以外,其余三个角为20°、40°、80°,有一定的规律。由此我想起一个三角恒等式: sinα·sin(60°-α)·sin(60°+α) =1/4sin3α(1) 如果在上题中令α=20°,则40°=60°-α,80°=60°+α,利用(1)式来解决就简单了。证:左=(3~(1/2))/2sin20°sin(60°-20°) ·sin(60°+20°) =(3~(1/2))/2·(1/4)sin60°=3/(16)=右。仿照(1)式,我们还可以证明
In the triangle, the proof of trigonometric function with product, simplification is a difficult point. For example, “Prove sin20 ° · sin40 ° · sin60 ° · sin80 ° = 3 / (16)”, generally need several times to apply the formula of integration and difference in order to prove. Careful observation of proof, the left except for the special angle of 60 °, the remaining three angles of 20 °, 40 °, 80 °, a certain law. So I think of a trigonometric identity: sinα · sin (60 ° -α) · sin (60 ° + α) = 1 / 4sin3α (1) 40 ° = 60 ° if α = 20 ° in the above α, 80 ° = 60 ° + α, using equation (1) to solve is simple. The left = (3 ~ (1/2)) / 2 sin 20 ° sin (60 ° -20 °) · sin (60 ° + 20 °) = (3 ~ (1/2)) / 2 · ) sin60 ° = 3 / (16) = right. Following the formula (1), we can prove it