Question

Given that 3sinA+5cosA=5, then find the value of (3cosA-5sinA)^2

Answer 1

Given that 3sinA+5cosA=5, then find the value of (3cosA-5sinA)^2

Solution,

3sinA+5cosA=5

By squaring both side,

(3sinA+5cosA)^2=25

(3sinA)^2+2.(3sinA).(5cosA)+(5cosA)^2=25

by using formula,

(a+b)^2=a^2+2ab+b^2

9sinA^2+ 2.3.5.sinA.cosA+25cosA^2=25

9sinA^2+30sinAcosA+25cosA^2=25

or, 9(1-cosA^2)+30sinAcosA+25(1-sinA^2)=25 { sinA^2=1-cosA^2 and cosA^2=1-sinA^2}

9-9cosA^2+30sinAcosA+25-25sinA^2-25=0

or, 9-9cosA²+30sinAcosA-25sinA²=0

or, -9+9cosA^2-30sinAcosA+25sinA^2=0

or, 9cosA^2-30sinAcosA+25sinA^2=9

or, (3cosA)^2-2.(3cosA).(5sinA^2)+(5sinA)^2=9

{ (a-b)^2=a^2-2ab+b^2}

(3cosA-5sinA)^2=9

3cosA-5sinA= (+/-)root(9)

3cosA-5sinA=+/-3