求证:2(cos θ -sin θ )/(1+sinθ +cosθ)=tan(∏/4- θ /2)-tan(θ /2)
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求证:2(cos θ -sin θ )/(1+sinθ +cosθ)=tan(∏/4- θ /2)-tan(θ /2)
看到右边是关于θ/2的,就想着把左边也化为θ/2的.
2(cos θ -sin θ )/(1+sinθ +cosθ)
上面=2((cosθ/2)^2-(sinθ/2)^2-
2(sinθ/2)(cosθ/2))
下面=(cosθ/2)^2+(sinθ/2)^2+
2(sinθ/2)(cosθ/2))
+(cosθ/2)^2-(sinθ/2)^2,
上下除以(cosθ/2)^2
得(1-tan(θ/2)^2-2tan(θ /2))除以
1+tan(θ/2)
而tan(∏/4- θ /2)-tan(θ /2)=
(1-tan(θ/2))/(1+tan(θ/2))+tan(θ/2)
通分得(1-tan(θ/2)^2-2tan(θ /2))除以
1+tan(θ/2)
即左边=右边
2(cos θ -sin θ )/(1+sinθ +cosθ)
上面=2((cosθ/2)^2-(sinθ/2)^2-
2(sinθ/2)(cosθ/2))
下面=(cosθ/2)^2+(sinθ/2)^2+
2(sinθ/2)(cosθ/2))
+(cosθ/2)^2-(sinθ/2)^2,
上下除以(cosθ/2)^2
得(1-tan(θ/2)^2-2tan(θ /2))除以
1+tan(θ/2)
而tan(∏/4- θ /2)-tan(θ /2)=
(1-tan(θ/2))/(1+tan(θ/2))+tan(θ/2)
通分得(1-tan(θ/2)^2-2tan(θ /2))除以
1+tan(θ/2)
即左边=右边
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