1-2sin30°cos30°
=1-sin60°
=1-√3/2
cos60°/(1+sin60°)+ 1/tan30°
=1/2/(1+√3/2 )+1/√3/3
=1/( 2+√3)+√3
=(4+2√3)/(2+√3)
=(4+2√3)(2-√3)/(2+√3)(2-√3)
=8-6
=2請問(1/2)/(1+√3/2 )怎么變成1/(2+√3)的，還有1/(√3/3 )怎么變√3的？(1/2)/(1+√3/2 )=(1/2)/(2/2+√3/2)=(1/2)/[(2+√3)/2]=(1/2)*[2/(2+√3)]=1/(2+√3)1/(√3/3 )=(1*3)/√3=√3*(1*3)/(√3*√3)=3*√3/3=√3看不懂這個1/( 2+√3)+√3變成這個(4+2√3)/(2+√3)1/( 2+√3)+√3=1/( 2+√3)+√3*( 2+√3)/( 2+√3)=(1+2√3+3)/( 2+√3)=(4+2√3)/(2+√3)