sinA + cosA =3 then find the value of sinA x cosA
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squaring on both sides
(sinA+cosA)^2=9
sin^2 A+cos^2 A+2sinAcosA=9
we know that sin^2A+cos^2A=1
then we get
1+2sinAcosA=9
2sinAcosA=9-1
sinAcosA=(8/2)
sinAcosA=4
(sinA+cosA)^2=9
sin^2 A+cos^2 A+2sinAcosA=9
we know that sin^2A+cos^2A=1
then we get
1+2sinAcosA=9
2sinAcosA=9-1
sinAcosA=(8/2)
sinAcosA=4
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