三角函数的求值和化简题!1.已知a为第二象限角,用Sina=√15/4(4分之根号15)来求Sin(a+π/4)/(Sin2a+Cos2a+1)2.求证Tan^X+1/Tan^X = (3+Cos4x)/(1-Cos4X)3.Sin9a=b)=5/13,tan(π/2)=1/2,a属于第一二象限,b属于任何一个象限,
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![三角函数的求值和化简题!1.已知a为第二象限角,用Sina=√15/4(4分之根号15)来求Sin(a+π/4)/(Sin2a+Cos2a+1)2.求证Tan^X+1/Tan^X = (3+Cos4x)/(1-Cos4X)3.Sin9a=b)=5/13,tan(π/2)=1/2,a属于第一二象限,b属于任何一个象限,](/uploads/image/z/8606493-45-3.jpg?t=%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0%E7%9A%84%E6%B1%82%E5%80%BC%E5%92%8C%E5%8C%96%E7%AE%80%E9%A2%98%211.%E5%B7%B2%E7%9F%A5a%E4%B8%BA%E7%AC%AC%E4%BA%8C%E8%B1%A1%E9%99%90%E8%A7%92%2C%E7%94%A8Sina%3D%E2%88%9A15%2F4%284%E5%88%86%E4%B9%8B%E6%A0%B9%E5%8F%B715%29%E6%9D%A5%E6%B1%82Sin%28a%2B%CF%80%2F4%29%2F%28Sin2a%2BCos2a%2B1%292.%E6%B1%82%E8%AF%81Tan%5EX%2B1%2FTan%5EX+%3D+%283%2BCos4x%29%2F%281-Cos4X%293.Sin9a%3Db%29%3D5%2F13%2Ctan%28%CF%80%2F2%29%3D1%2F2%2Ca%E5%B1%9E%E4%BA%8E%E7%AC%AC%E4%B8%80%E4%BA%8C%E8%B1%A1%E9%99%90%2Cb%E5%B1%9E%E4%BA%8E%E4%BB%BB%E4%BD%95%E4%B8%80%E4%B8%AA%E8%B1%A1%E9%99%90%2C)
三角函数的求值和化简题!1.已知a为第二象限角,用Sina=√15/4(4分之根号15)来求Sin(a+π/4)/(Sin2a+Cos2a+1)2.求证Tan^X+1/Tan^X = (3+Cos4x)/(1-Cos4X)3.Sin9a=b)=5/13,tan(π/2)=1/2,a属于第一二象限,b属于任何一个象限,
三角函数的求值和化简题!
1.已知a为第二象限角,用Sina=√15/4(4分之根号15)来求Sin(a+π/4)/(Sin2a+Cos2a+1)
2.求证Tan^X+1/Tan^X = (3+Cos4x)/(1-Cos4X)
3.Sin9a=b)=5/13,tan(π/2)=1/2,a属于第一二象限,b属于任何一个象限,求Sinb 和 cosb 和cosa
有一题打错了...
3..Sin(a+b)=5/13,tan(π/2)=1/2,a属于第一二象限,b属于任何一个象限,求Sinb 和 cosb 和cosa
三角函数的求值和化简题!1.已知a为第二象限角,用Sina=√15/4(4分之根号15)来求Sin(a+π/4)/(Sin2a+Cos2a+1)2.求证Tan^X+1/Tan^X = (3+Cos4x)/(1-Cos4X)3.Sin9a=b)=5/13,tan(π/2)=1/2,a属于第一二象限,b属于任何一个象限,
1,Sin(a+π/4)/(Sin2a+Cos2a+1)= [2分之根号2*(sina+cosa)]/(sin2a+2cos平方a)=[2分之根号2*(sina+cosa)]/(2sinacosa+2cos平方 a)=[2分之根号2*(sina+cosa)]/[2cosa*(sina+cosa)]=2分之根号2/(cosa*2)
由已知条件得cosa=-1/4带入得 :负根号2
2,=tan²x+1/tan²x
=sin²x/cos²x+cos²x/sin²x
=[(sinx)^4+(cosx)^4]/(sinxcosx)²
={[(sinx)²+(cosx)²]²-2(sinxcosx)²}/(sinxcosx)²
=[1-2(sinxcosx)²]/(sinxcosx)²
=4[1-1/2(sin2x)²]/(sin2x)²
=2[4-2(sin2x)²]/(1-cos4x)
=2(3+cos4x)/(1-cos4x)
3,tanb=2tan(b/2)/(1-tan平方b/2)=4/3
2
证:
=tan²x+1/tan²x
=sin²x/cos²x+cos²x/sin²x
=[(sinx)^4+(cosx)^4]/(sinxcosx)²
={[(sinx)²+(cosx)²]²-2(sinxcosx)²}/(sinxcosx)...
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2
证:
=tan²x+1/tan²x
=sin²x/cos²x+cos²x/sin²x
=[(sinx)^4+(cosx)^4]/(sinxcosx)²
={[(sinx)²+(cosx)²]²-2(sinxcosx)²}/(sinxcosx)²
=[1-2(sinxcosx)²]/(sinxcosx)²
=4[1-1/2(sin2x)²]/(sin2x)²
=2[4-2(sin2x)²]/(1-cos4x)
=2(3+cos4x)/(1-cos4x)
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