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Trigonometry> efbs

question mark

efbs

shreyas , 10 Years ago

Grade

1 Answers

Jitender Singh

Last Activity: 10 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

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