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

Jitender Singh IIT Delhi
9 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
Sunil Kumar FP
9 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