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How to prove this identity: 1-sin2x/cosx = cosx?
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do I expand sin2x? eep.
is that sin (2x) or sin^2 x?
sin^2x
lemme rephrase that.. \[\frac{1 - \sin^2x}{\cos^2x}\]
yup
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i mean cos x
okay
change \(1 - \sin^2x\) into one of the good old trig identities
is that sin2x or sin^2x
cos^2x?
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got it! \[\frac{\cos^2 x}{\cos x}\] simplify that
lol mimi :P
\(sin^2x+cos^2x=1\) => \(cos^2x=1-sin^2x\)
ohhh, i cancel: cosxcosx/cosx?
\[\frac{\cancel{(\cos x)}(\cos x)}{\cancel{\cos x}}\]
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