1.在△ABC中,AB=AC,且sinB=8/17,求sinA2.已知sina=3/5,切a∈(90°,180°),求cos(2a-60°)3.已知tana,tanb是方程2x平方+x-6=0的两个根,求tan(a+b)的值4.证明:(cosa+sina)(cosa-sina)/2sinacosa=cot2a5.tan81°-cot54°/1+tan81°cot54°=6.若
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![1.在△ABC中,AB=AC,且sinB=8/17,求sinA2.已知sina=3/5,切a∈(90°,180°),求cos(2a-60°)3.已知tana,tanb是方程2x平方+x-6=0的两个根,求tan(a+b)的值4.证明:(cosa+sina)(cosa-sina)/2sinacosa=cot2a5.tan81°-cot54°/1+tan81°cot54°=6.若](/uploads/image/z/6557925-21-5.jpg?t=1.%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2CAB%3DAC%2C%E4%B8%94sinB%3D8%2F17%2C%E6%B1%82sinA2.%E5%B7%B2%E7%9F%A5sina%3D3%2F5%2C%E5%88%87a%E2%88%88%2890%C2%B0%2C180%C2%B0%29%2C%E6%B1%82cos%282a-60%C2%B0%293.%E5%B7%B2%E7%9F%A5tana%2Ctanb%E6%98%AF%E6%96%B9%E7%A8%8B2x%E5%B9%B3%E6%96%B9%2Bx-6%3D0%E7%9A%84%E4%B8%A4%E4%B8%AA%E6%A0%B9%2C%E6%B1%82tan%28a%2Bb%29%E7%9A%84%E5%80%BC4.%E8%AF%81%E6%98%8E%3A%28cosa%2Bsina%29%28cosa-sina%29%2F2sinacosa%3Dcot2a5.tan81%C2%B0-cot54%C2%B0%2F1%2Btan81%C2%B0cot54%C2%B0%3D6.%E8%8B%A5)
1.在△ABC中,AB=AC,且sinB=8/17,求sinA2.已知sina=3/5,切a∈(90°,180°),求cos(2a-60°)3.已知tana,tanb是方程2x平方+x-6=0的两个根,求tan(a+b)的值4.证明:(cosa+sina)(cosa-sina)/2sinacosa=cot2a5.tan81°-cot54°/1+tan81°cot54°=6.若
1.在△ABC中,AB=AC,且sinB=8/17,求sinA
2.已知sina=3/5,切a∈(90°,180°),求cos(2a-60°)
3.已知tana,tanb是方程2x平方+x-6=0的两个根,求tan(a+b)的值
4.证明:(cosa+sina)(cosa-sina)/2sinacosa=cot2a
5.tan81°-cot54°/1+tan81°cot54°=
6.若cosa=1/5,且3/2∏
(要求写出解答过程)答一题得一题`谢谢了
8.若sina=1/根号5,sinb=1/根号10,且a,b∈(0,90°),则a+b=
9.若sin x/2-cos x/2=1/3,则sinx=
10.在△ABC中,已知sinA=4/5,cosB=5/13,那么cosC=
1.在△ABC中,AB=AC,且sinB=8/17,求sinA2.已知sina=3/5,切a∈(90°,180°),求cos(2a-60°)3.已知tana,tanb是方程2x平方+x-6=0的两个根,求tan(a+b)的值4.证明:(cosa+sina)(cosa-sina)/2sinacosa=cot2a5.tan81°-cot54°/1+tan81°cot54°=6.若
第一题需要画图,你可以根据题意画个等腰三角形ABC,过点A向BC作垂线,垂足为点D,则三角形ABO全等于三角形ACO(易证明),再过点B向AC作垂线,垂足为点E,则三角形BCE相似于三角形ABO(易证明) 根据勾股定理可得BO=15,则BC=30,可列AO/AB=8/17=BE/BC,易得BE为:240/17 sinA=BE/AB,把AB=17,BE=240/17代入BE/AB中得sinA=240/289 注:(AB=17 AO=8为假设)第3题先解方程2x平方+x-6=0,可得.