Question

已知a是第一象限角，且cosa=3/5，那么[1+√2cos(2a-pi/4)]/sin(a+pi/2)=

Answer 1

cosa=3/5,a是第一象限角=>sina=4/5
[1+√2cos(2a-pi/4)]/sin(a+pi/2)= [1+√2(cos2acospi/4+sin2asinpi/4)]/cosa
=[1+cos2a+sin2a]/cosa (因为：cospi/4=sinpi/4=√2/2)
=[1+(2cosa*cosa-1)+2sina*cosa]/cosa
=2cosa(cosa+sina)/cosa
=2(cosa+sina)
=2(3/5+4/5)
=14/5

Answer 2

a为一象限角，故cosa=4/5
化简 √2cos(2a-π/4)=√2[cos2acosπ/4+sin2asinπ/4]=2(cos2a+sin2a)=2(cosacosa-sinasina+2sinacosa)=2(16/25-9/25+2×3/5×4/5)=62/25
1+62/25=87/25
sin(a+π/2)=sina=3/5
87/25÷3/5=29/5

Answer 3

sin(a+π/2)=cosa=3/5,已知a是第一象限角，且cosa=3/5，sina=4/5,
原式=(1+cos2a+sin2a)/cosa=(2cos^2a+2sinacosa)/cosa=2(cosa+sina)=14/5

Answer 4

cosa=3/5,a是第一象限角=>sina=4/5
[1+√2cos(2a-pi/4)]/sin(a+pi/2)=
[1+√2(cos2acospi/4+sin2asinpi/4)]/cosa
=[1+cos2a+sin2a]/cosa
(因为：cospi/4=sinpi/4=√2/2)
=[1+(2cosa*cosa-1)+2sina*cosa]/cosa
=2cosa(cosa+sina)/cosa
=2(cosa+sina)
=2(3/5+4/5)
=14/5