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Trigonometry

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

shreyas

12 Years agoGrade

2 Answers

Jitender Singh

12 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

12 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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