设f(a)=2sin(π+a)cos(π-a)-cos(π+a)/1+sin^2a+cos(3π/2+a)-sin^2(π/2+a),且1+2sina≠0,化简f(a)
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![设f(a)=2sin(π+a)cos(π-a)-cos(π+a)/1+sin^2a+cos(3π/2+a)-sin^2(π/2+a),且1+2sina≠0,化简f(a)](/uploads/image/z/665950-22-0.jpg?t=%E8%AE%BEf%28a%29%3D2sin%28%CF%80%2Ba%29cos%28%CF%80-a%29-cos%28%CF%80%2Ba%29%2F1%2Bsin%5E2a%2Bcos%283%CF%80%2F2%2Ba%29-sin%5E2%28%CF%80%2F2%2Ba%29%2C%E4%B8%941%2B2sina%E2%89%A00%2C%E5%8C%96%E7%AE%80f%28a%29)
设f(a)=2sin(π+a)cos(π-a)-cos(π+a)/1+sin^2a+cos(3π/2+a)-sin^2(π/2+a),且1+2sina≠0,化简f(a)
设f(a)=2sin(π+a)cos(π-a)-cos(π+a)/1+sin^2a+cos(3π/2+a)-sin^2(π/2+a),且1+2sina≠0,化简f(a)
设f(a)=2sin(π+a)cos(π-a)-cos(π+a)/1+sin^2a+cos(3π/2+a)-sin^2(π/2+a),且1+2sina≠0,化简f(a)
1.f(a)=[2sin( π+a)cos(π-a)-cos(π+a)]/[1+sin^2a+sin(π-a)-cos^2(π-a)]
=(2sinacosa+cosa)/(1+sin²a+sina-cos²a)
=cosa(2sina+1)/[sina(2sina+1)]
=cota
(1) a=-17π/6
f(a)=cot(-17π/6)=-cot(17π/6)=-cot(3π-π/6)
=cot(π/6)=√3
(2) 若a是锐角,且sin(a-3π/2)=3/5
即sin(3π/2-a)=-3/5
-cosa=-3/5 cosa=3/5
sina=√(1-cos²a)=4/5
所以f(a)=cota=cosa/sina=(3/5)/(4/5)=3/4
2.已知sin&=3/5,&属于(π/2,π),tan&