cos(π/4+x)=5/13.且0
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cos(π/4+x)=5/13.且0 cos(π/4+x)=5/13.且0 ∵0 全部展开 ∵0 收起 cos(π/4+x)=sin(π/2-π/4-x)=sin(π/4-x)=5/13=-sin(x-π/4); 全部展开 cos(π/4+x)=sin(π/2-π/4-x)=sin(π/4-x)=5/13=-sin(x-π/4); 收起 sin(π/4 + x)=±√1 - cos²(π/4 + x)=±√1 - (5/13)²=±12/13 全部展开 sin(π/4 + x)=±√1 - cos²(π/4 + x)=±√1 - (5/13)²=±12/13 收起
cos(π/4+x)=5/13.且0
1.设a=π/4+x 则cosa=5/13 且π/4
又∵cos(π/4+x)=5/13
∴sin(π/4+x)=12/13
∴cos2x/sin(π/4 -x)
=sin(2x+π/2)/cos(π/4-x-π/2)
=2sin(x+π/4)cos(x+π/4)/cos(-π/4-x)
=2sin(x+π/4)cos(x+π/4)/cos(π/4...
又∵cos(π/4+x)=5/13
∴sin(π/4+x)=12/13
∴cos2x/sin(π/4 -x)
=sin(2x+π/2)/cos(π/4-x-π/2)
=2sin(x+π/4)cos(x+π/4)/cos(-π/4-x)
=2sin(x+π/4)cos(x+π/4)/cos(π/4+x)
=2sin(x+π/4)
=24/13
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∵0
∴sin(π/4+x)=√(1-25/169)=12/13=cos(π/4-x)=cos(x-π/4);
∴cos2x=cos(x+π/4+x-π/4)=cos(x+π/4)cos(x-π/4)-sin(x+π...
∵0
∴sin(π/4+x)=√(1-25/169)=12/13=cos(π/4-x)=cos(x-π/4);
∴cos2x=cos(x+π/4+x-π/4)=cos(x+π/4)cos(x-π/4)-sin(x+π/4)sin(x-π/4)=(5/13)×(12/13)-(12/13)×(-5/13)=120/169;
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∵0＜x＜π/4
∴π/4＜π/4 + x＜π/2
∴sin(π/4 + x)=12/13
cos2x/sin(π/4 - x)=(cos²x - sin²x)/[sin(π/4)cosx - cos(π/4)sinx]...
∵0＜x＜π/4
∴π/4＜π/4 + x＜π/2
∴sin(π/4 + x)=12/13
cos2x/sin(π/4 - x)=(cos²x - sin²x)/[sin(π/4)cosx - cos(π/4)sinx]
=[(cosx+sinx)(cosx-sinx)]/[(√2/2)(cosx-sinx)]
=(√2)(cosx+sinx)
=2[(√2/2)(cosx+sinx)]
=2[(√2/2)cosx + (√2/2)sinx]
=2sin(π/4 + x)
=2 × 12/13
=24/13