An alternating voltage is given by e=(6sinωt+8cosωt) volt. The peak value of voltage is given by:
e=6sinωt+8cosωt
⟹e=6cos(ωt−90)+8cosωt
Taking maximum values
⟹e=6∠−90+8∠0
⟹e=6cos90−6jsin90+8
⟹e=8−6j
⟹e=10∠−36.86
⟹e=10cos(ωt−36.86)
So, maximum or peak value =10