Asked by bhavya
cos@-4SIN@=1 then sin@+4cos@=
Answers
Answered by
Reiny
From the first:
cosØ - 1 = 4sinØ
square both sides
cos^2 Ø - 2cosØ + 1 = 16sin^2 Ø
cos^2 Ø - 2cosØ + 1 = 16(1 - cos^2 Ø)
17cos^2 Ø - 2cosØ - 15 = 0
(17cosØ + 15)(cosØ - 1)
cosØ = -15/17 or cosØ = 1
case1: cosØ = -15/17
then sinØ = 8/17 (in 2nd quad)
then sinØ + 4cosØ = 8/17 + 4(-15)/17 = -52/17
or
sinØ = -8/17 (in 3rd quad)
then sinØ + 4cosØ = -8/17 + 4(-15/17) = -68/17
Case 2: cosØ = 1
then sinØ = 0
and sinØ + 4cosØ = 0 + 4(1) = 4
cosØ - 1 = 4sinØ
square both sides
cos^2 Ø - 2cosØ + 1 = 16sin^2 Ø
cos^2 Ø - 2cosØ + 1 = 16(1 - cos^2 Ø)
17cos^2 Ø - 2cosØ - 15 = 0
(17cosØ + 15)(cosØ - 1)
cosØ = -15/17 or cosØ = 1
case1: cosØ = -15/17
then sinØ = 8/17 (in 2nd quad)
then sinØ + 4cosØ = 8/17 + 4(-15)/17 = -52/17
or
sinØ = -8/17 (in 3rd quad)
then sinØ + 4cosØ = -8/17 + 4(-15/17) = -68/17
Case 2: cosØ = 1
then sinØ = 0
and sinØ + 4cosØ = 0 + 4(1) = 4
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