高一三角函数的定义和关系
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高一三角函数的定义和关系
1.计算
(1)7cos270°+12sin0°+8cos360°
(2)sin4次方乘4分之π-cos2次方乘2分之π+6tan3次方乘4分之π
2.已知tanα=3
求(1) (根号3cosα-sinα)除(根号3cosα+sinα)
(2)2sin2次方α-3sinα乘cosα
3.求证
(1) (1-2sinXcosX)除(cos2次方X-sin2次方X)=(1-tanX)除(1+tanX)
(2)sin4次方X+cos4次方X=2sin2次方Xcos2次方X
4.化简
(1) (1+tan2次方α)cos2次方α
(2)tanα-cotα-(1-2cos2次方α)除(sinα乘cosα)
1.计算
(1)7cos270°+12sin0°+8cos360°
(2)sin4次方乘4分之π-cos2次方乘2分之π+6tan3次方乘4分之π
2.已知tanα=3
求(1) (根号3cosα-sinα)除(根号3cosα+sinα)
(2)2sin2次方α-3sinα乘cosα
3.求证
(1) (1-2sinXcosX)除(cos2次方X-sin2次方X)=(1-tanX)除(1+tanX)
(2)sin4次方X+cos4次方X=2sin2次方Xcos2次方X
4.化简
(1) (1+tan2次方α)cos2次方α
(2)tanα-cotα-(1-2cos2次方α)除(sinα乘cosα)
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1.(1)7cos270°+12sin0°+8cos360°
=7*0+12*0+8*1
=8
(2)sin4次方乘4分之π-cos2次方乘2分之π+6tan3次方乘4分之π
=(√2/2)^4-0^2+6*1^3
=25/4
2.已知tanα=3
sina/cosa=3
sina=3cosa
求(1) (根号3cosα-sinα)除(根号3cosα+sinα)
=(√3cosa-3cosa)/(√3cosa+3cosa)
=(√3-3)/(√3+3)
=(12-6√3)/(-6)
=√3-2
(sina)^2+(cosa)^2=1
sina=3cosa
所以(cosa)^2=1/10
(2)2sin2次方α-3sinα乘cosα
=2*(3cos)^2-3*3cosacosa
=2*3^2*1/10-3*3*1/10
=9/10
3.求证
(1)(1-2sinxcosx)/(cos^2x-sin^2x)
=(cosx-sinx)^2/(cosx-sinx)(cosx+sinx)
=(cosx-sinx)/(cosx+sinx)
(1-tanx)/(1+tanx)
=(1-sinx/cosx)/(1+sinx/cosx)
=(cosx-sinx)/(cosx+sinx)
所以(1-2sinxcosx)/(cos^2x-sin^2x)=(1-tanx)/(1+tanx)
(2)sin4次方X+cos4次方X=2sin2次方Xcos2次方X
题目不对
4.化简
(1) (1+tan2次方α)cos2次方α
=[1+(sina)^2/(cosa)^2](cosa)^2
=(cosa)^2+(sina)^2
=1
(2)tanα-cotα-(1-2cos2次方α)除(sinα乘cosα)
=sina/cosa-cosa/sina-[1-2(cosa)^2]/(sinα*cosα)
=[(sina)^2-(cosa)^2-1+2(cosa)^2]/(sinα*cosα)
=[(sina)^2+(cosa)^2-1]/(sinα*cosα)
=(1-1)/(sinα*cosα)
=0
=7*0+12*0+8*1
=8
(2)sin4次方乘4分之π-cos2次方乘2分之π+6tan3次方乘4分之π
=(√2/2)^4-0^2+6*1^3
=25/4
2.已知tanα=3
sina/cosa=3
sina=3cosa
求(1) (根号3cosα-sinα)除(根号3cosα+sinα)
=(√3cosa-3cosa)/(√3cosa+3cosa)
=(√3-3)/(√3+3)
=(12-6√3)/(-6)
=√3-2
(sina)^2+(cosa)^2=1
sina=3cosa
所以(cosa)^2=1/10
(2)2sin2次方α-3sinα乘cosα
=2*(3cos)^2-3*3cosacosa
=2*3^2*1/10-3*3*1/10
=9/10
3.求证
(1)(1-2sinxcosx)/(cos^2x-sin^2x)
=(cosx-sinx)^2/(cosx-sinx)(cosx+sinx)
=(cosx-sinx)/(cosx+sinx)
(1-tanx)/(1+tanx)
=(1-sinx/cosx)/(1+sinx/cosx)
=(cosx-sinx)/(cosx+sinx)
所以(1-2sinxcosx)/(cos^2x-sin^2x)=(1-tanx)/(1+tanx)
(2)sin4次方X+cos4次方X=2sin2次方Xcos2次方X
题目不对
4.化简
(1) (1+tan2次方α)cos2次方α
=[1+(sina)^2/(cosa)^2](cosa)^2
=(cosa)^2+(sina)^2
=1
(2)tanα-cotα-(1-2cos2次方α)除(sinα乘cosα)
=sina/cosa-cosa/sina-[1-2(cosa)^2]/(sinα*cosα)
=[(sina)^2-(cosa)^2-1+2(cosa)^2]/(sinα*cosα)
=[(sina)^2+(cosa)^2-1]/(sinα*cosα)
=(1-1)/(sinα*cosα)
=0