高一的数学问题、

已知角β在第一象限且cosa=3/5.则1+ √2cos（2a- π/4）除以 sin（a+π/2）=？
sina=4/5
(1+(2^(1/2))*(cos2acosπ/4+sin2asinπ/4))/(-cosa)
=(1+（2cosacosa-1+2sinacosa))/(-cosa)=-2(cosa+sina)=-2(3/5+4/5)=-14/5

b在第一象限且cosb=3/5.

(1+√2cos(2b-45))/sin(b+45)
=√2(1+cos2b+sin2b)/(sinb+cosb)
=√2[(sinb+cosb)^2+(sincosb)(cosb-sinb)]/(sinb+cosb)
=√2(sinb+cosb)(2cosb)/(sinb+cosb)
=2√2cosb
=6√2/5