tan(a+b)= n tan(a-b) n≠-1 a≠kπ/2 求证 sin2b/sin2a = n-1/n+1
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tan(a+b)= n tan(a-b) n≠-1 a≠kπ/2 求证 sin2b/sin2a = n-1/n+1
sin2b/sin2a=sin((a+b)-(a-b))/sin((a+b)+(a-b))
=[sin(a+b)cos(a-b)-sin(a-b)cos(a+b)]/[sin(a+b)cos(a-b)+sin(a-b)cos(a+b)](分子分母同时除以cos(a-b)cos(a+b))
=[tan(a+b)-tan(a-b)]/[tan(a+b)+tan(a-b)]
=[n tan(a-b)-tan(a-b)]/[n tan(a-b)+tan(a-b)]
=n-1/n+1
=[sin(a+b)cos(a-b)-sin(a-b)cos(a+b)]/[sin(a+b)cos(a-b)+sin(a-b)cos(a+b)](分子分母同时除以cos(a-b)cos(a+b))
=[tan(a+b)-tan(a-b)]/[tan(a+b)+tan(a-b)]
=[n tan(a-b)-tan(a-b)]/[n tan(a-b)+tan(a-b)]
=n-1/n+1
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