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Trigonometry

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Profile image of shreyas
12 Years agoGrade
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1 Answer

Profile image of 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

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