please find the least valueof cos^2-6sinacosa+2+3sin^2a
45 , 5 Years ago
Grade 11
Trigonometry> please find the least value of cos^2-6sin...
2 AnswersVikas TU
Last Activity: 5 Years ago
Dear student
Cos^2Q - 6SinQ*CosQ + 3Sin^2Q + 2
Cos^2Q + Sin^2Q +1 - Cos2Q -3Sin2Q
=> 1+1 - Cos2Q -3Sin2Q +2
=> 4 - [Cos2Q-3sin2Q]
Minimum value
=> 4 - sqrt((-1)^2 + (-3)^2 )
=> 4- sqrt(10)
Cos^2Q + Sin^2Q +1 - Cos2Q -3Sin2Q
=> 1+1 - Cos2Q -3Sin2Q +2
=> 4 - [Cos2Q-3sin2Q]
Minimum value
=> 4 - sqrt((-1)^2 + (-3)^2 )
=> 4- sqrt(10)
Hope this helps
Arun
Last Activity: 5 Years ago
maximum value is 3
cos^2a-6sinacosa +2+3sin^2a=cos^2a +sin^2a +2sin^2a -6sina cosa=3 +2sin2a(sin2a-3)
it will have maximum value when second term is maximum value at sin2a minimum is -1
therefore 3+2*-1(-1-3)=11
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