高数,尽快!

6.已知tanα=3,求sinα,cosα.
tanα=3,cotα=1/3,csc²α=1+cot²α=1+(1/3)²=10/9,故cscα=±(1/3)√10,sinα=±(3/10)√10；
sec²α=1+tan²α=1+3²=10,故secα=±√10,cosα=±1/√10=±(1/10)√10.
7.已知cos(π/6-α)=1/3,求cos(11π/6-α)的值.
由cos(π/6-α)=1/3,得π/6-α=arccos(1/3),故α=π/6-arccos(1/3);
于是cosα=cos[π/6-arccos(1/3)]=cos(π/6)cos[arccos(1/3)]-sin(π/6)sin[arccos(1/3)]
=(√3/2)(1/3)-(1/2)√(1-1/9)=√3/6-2√2/6=(√3-2√2)/6；
cos(11π/6-α)=cos[2π-(π/6+α)]=cos(π/6+α)；
cos(π/6+α)+cos(π/6-α)=2cos(π/6)cosα=(√3)cosα=(√3)(√3-2√2)/6=(3-2√6)/6；
∴cos(11π/6-α)=cos(π/6+α)=(3-2√6)/6-cos(π/6-α)=(3-2√6)/6-(1/3)=(1-2√6)/6.