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maxima of cos^2-6sinacosa+2+3sin^2a

maxima of cos^2-6sinacosa+2+3sin^2a

Grade:

2 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
6 years ago
Ans:
f(a) = cos^{2}a - 6sina.cosa + 3sin^{2}a+2
f(a) = cos^{2}a+sin^{2}a-sin^{2}a - 6sina.cosa + 3sin^{2}a+2
f(a) = 2sin^{2}a - 6sina.cosa + 3
f(a) = 2sin^{2}a - 3sin2a + 3
f'(a) = 4sina.cosa - 3.2cos2a
f'(a) = 2sin2a- 6cos2a
For maxima:
f'(a) = 0
2sin2a- 6cos2a =0
tan2a = 3
\Rightarrow sin2a = \pm \frac{3}{\sqrt{10}}, cos2a = \pm \frac{1}{\sqrt{10}}
f''(a) = 4cos2a + 12sin2a
f''(a) < 0
when
\Rightarrow sin2a = - \frac{3}{\sqrt{10}}, cos2a = - \frac{1}{\sqrt{10}}
f(a) = 2sin^{2}a - 3sin2a + 3
f(a) = 2sin^{2}a-1 - 3sin2a + 4
f(a) = -cos2a - 3sin2a + 4
Maximum value:
f(a) = -(\frac{-1}{\sqrt{10}}) - 3(\frac{-3}{\sqrt{10}}) + 4
f(a) =\sqrt{10} + 4
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty
Sunil Kumar FP
askIITians Faculty 183 Points
6 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
thanks and regards
sunil kr
askIITian faculty

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