已知tana=1/2求下列各式的值 sin^2a+sina*cosa+2

网友回答

tan(a)=1/2
sin²(a)+sin(a)*cos(a)+2
=cos²(a)*[tan²(a)+tan(a)]+2
=cos²(a)*(3/4)+2
=1/[1/cos²(a)]*(3/4)+2
=1/{[sin²(a)+cos²(a)]/cos²(a)}*(3/4)+2
=1/[tan²(a)+1]*(3/4)+2
=1/(1/4+1)*(3/4)+2
=(4/5)*(3/4)+2
=3/5+2
=13/5======以下答案可供参考======
供参考答案1:
解:用万能公式:sin2a=2tana/(1+tan^2a)=1/(1+1/4)=4/5;
cos2a=(1-tan^2a)/(1+tan^2a)=(1-1/4)/(1+1/4)=3/5.
所以,原式=(1-cos2a)/2+sin2a/2+2
=1/5+2/5+2=13/5