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Find the absolute minimum value of f(x)=2sin^2(x) + cos^2(x) on [-pi,pi]
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I'm still tempted to do this: \[f(x) = 2\sin^{2}(x) + \cos^{2}(x) = \sin^{2}(x) + \sin^{2}(x) + \cos^{2}(x) = \sin^{2}(x) + (\sin^{2}(x) + \cos^{2}(x)) = \sin^{2}(x) + 1\] then it seems kind of obvious.
In case you can't see that, since I don't feel like typing it all over, split off one (1) of the sines and combine it with the cosine. This is then the Pythagorean Identity and it can be replaced by "1". \[\sin^{2}(x) + 1\] is much easier to see.
thankyou i got it :]
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