Question

这几道数学题怎么解？

1:求值：tan70°×cos10°（根号3×tan20°—1）
2：已知cosA=1/7,cos(A-B)=13/14,且0＜B＜A＜2/π，求B的大小（A、B表示角）
3：已知函数f(x)=2cos2x+sin²x-4cosx。（1）求f(π/3)的值；（2）求f(x)的最大值和最小值
4：设f(x)=6cos²x-根号3(sin2x)。（1）求f(x)的最大值及最小值正周期；（2）若锐角a满足f(a)=3-2倍根号3，求tan4/5(a)的值。
谢谢~！
答得快、好还会加分！
哪怕只做得出一两道题都行！

Answer 1

tan70×cos10×(√3tan20-1)
=tan70×cos10×(tan60×tan20-1)
=tan70×cos10×[(sin60×sin20/cos60×cos20)-1]
=tan70×cos10×(sin60×sin20-cos60×cos20)/(cos60×cos20)
=tan70×cos10×[-cos(60+20)]/(cos60 ×cos20)
=-tan70×cos10×cos80/(cos60×cos20)
=-tan70×cos10×sin10/(cos60×cos20)
=-(sin70/cos70)×(1/2)×sin20/(cos60×cos20)
=-(cos20/sin20)×sin20/(2cos60×cos20)
=-1/(2cos60)
=-1
cos a=1/7 且0＜a＜π/2故sin a>0且
sin²a=1-cos²=1-1/49=48/49故sin a=4√3/7
而cos(a-b)=cos a cos b+sin a sin b=1/7cos b+4√3/7sin b=13/14即
cos b+4√3sin b=13/2 ①
由0＜a-b＜π/2知 sin (a-b)>0 而sin²(a-b)=1-cos²(a-b)=1-169/196=27/196
sin(a-b)=3√3/14=sin a cos b-cos a sin b=4√3/7cos b-1/7sin b即
4√3cos b-sin b=3√3/2 两边同时乘以4√3,有
48cos b-4√3sin b=18 ②
两式相加，有
49cos b=49/2则cos b=1/2 由0＜b＜π/2知b =arc cos 1/2=π/3

f（x）=2cos2x+sin²x-4cosx =2cos(2π/3)+sin²(π/3)-4cos(π/3)
=2*(-1/2)+3/4-4*1/2
=-9/4
f（x）=2cos2x+sin²x-4cosx=2(2cos²x-1)+(1-cos²x)-4cosx=3cos²x)-4cosx-1
=3(cosx-2/3)²-7/3
当cosx=1时，最小值为-2

f(x)=6cos^2x-√3sin2x
=3(1+cos2x)-√3sin2x
=3+2√3(cosπ/6cos2x-sinπ/6sin2x)
=3+2√3cos(2x+π/6)
fmax(x)=3+2√3
fmin(x)=3-2√3
2）
f(a)=fmin(x)
2a+π/6=π,a=π/2-π/12=5π/12
4a/5=π/3
tan4/5a=tanπ/3=√3
当cosx=-1时，最大值为6

Answer 2

tan70×cos10×(√3tan20-1)
=tan70×cos10×(tan60×tan20-1)
=tan70×cos10×[(sin60×sin20/cos60×cos20)-1]
=tan70×cos10×(sin60×sin20-cos60×cos20)/(cos60×cos20)
=tan70×cos10×[-cos(60+20)]/(cos60 ×cos20)
=-tan70×cos10×cos80/(cos60×cos20)
=-tan70×cos10×sin10/(cos60×cos20)
=-(sin70/cos70)×(1/2)×sin20/(cos60×cos20)
=-(cos20/sin20)×sin20/(2cos60×cos20)
=-1/(2cos60)
=-1

Answer 3

tan70×cos10×(√3tan20-1)
=tan70×cos10×(tan60×tan20-1)
=tan70×cos10×[(sin60×sin20/cos60×cos20)-1]
=tan70×cos10×(sin60×sin20-cos60×cos20)/(cos60×cos20)
=tan70×cos10×[-cos(60+20)]/(cos60 ×cos20)
=-tan70×cos10×cos80/(cos60×cos20)
=-tan70×cos10×sin10/(cos60×cos20)
=-(sin70/cos70)×(1/2)×sin20/(cos60×cos20)
=-(cos20/sin20)×sin20/(2cos60×cos20)
=-1
cos a=1/7 且0＜a＜π/2故sin a>0且
sin²a=1-cos²=1-1/49=48/49故sin a=4√3/7
而cos(a-b)=cos a cos b+sin a sin b=1/7cos b+4√3/7sin b=13/14即
cos b+4√3sin b=13/2 ①
由0＜a-b＜π/2知 sin (a-b)>0 而sin²(a-b)=1-cos²(a-b)=1-169/196=27/196
sin(a-b)=3√3/14=sin a cos b-cos a sin b=4√3/7cos b-1/7sin b即
4√3cos b-sin b=3√3/2 两边同时乘以4√3,有
48cos b-4√3sin b=18 ②
两式相加，有
49cos b=49/2则cos b=1/2 由0＜b＜π/2知b =arc cos 1/2=π/3

f（x）=2cos2x+sin²x-4cosx =2cos(2π/3)+sin²(π/3)-4cos(π/3)
=2*(-1/2)+3/4-4*1/2
=-9/4
f（x）=2cos2x+sin²x-4cosx=2(2cos²x-1)+(1-cos²x)-4cosx=3cos²x)-4cosx-1
=3(cosx-2/3)²-7/3
当cosx=1时，最小值为-2

f(x)=6cos^2x-√3sin2x
=3(1+cos2x)-√3sin2x
=3+2√3(cosπ/6cos2x-sinπ/6sin2x)
=3+2√3cos(2x+π/6)
fmax(x)=3+2√3
fmin(x)=3-2√3
2）
f(a)=fmin(x)
2a+π/6=π,a=π/2-π/12=5π/12
4a/5=π/3
tan4/5a=tanπ/3=√3
当cosx=-1时，最大值为6

Answer 4

tan70×cos10×(√3tan20-1)
=tan70×cos10×(tan60×tan20-1)
=tan70×cos10×[(sin60×sin20/cos60×cos20)-1]
=tan70×cos10×(sin60×sin20-cos60×cos20)/(cos60×cos20)
=tan70×cos10×[-cos(60+20)]/(cos60 ×cos20)
=-tan70×cos10×cos80/(cos60×cos20)
=-tan70×cos10×sin10/(cos60×cos20)
=-(sin70/cos70)×(1/2)×sin20/(cos60×cos20)
=-(cos20/sin20)×sin20/(2cos60×cos20)
=-1/(2cos60)
=-1
其它的我不会