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given A=sin^2a+cos^4a,then for value of a,(a=theta)

6 years ago


Answers : (1)

										Ans:Hello student, please find answer to your question
A = sin^{2}a + cos^{4}a
A = sin^{2}a + cos^{2}a.cos^{2}a
A = sin^{2}a + cos^{2}a(1-sin^{2}a)
A = sin^{2}a + cos^{2}a-sin^{2}a.cos^{2}a
A = 1-sin^{2}a.cos^{2}a
A = 1-\frac{4}{4}.sin^{2}a.cos^{2}a
4A = 4-(2sina.cosa)^{2}
4A = 4-(sin2a)^{2}
(sin2a)^{2} = 4-4A
(sin2a)^{2} = 4(1-A)
sin2a = \pm2\sqrt{1-A}
sin^{-1}sin2a = sin^{-1}(\pm2\sqrt{1-A})
2a = sin^{-1}(\pm2\sqrt{1-A})
a =\frac{1 }{2}.sin^{-1}(\pm2\sqrt{1-A})
2 years ago

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