I am just trying to figure out why this happens: $$ \cos x + \frac{\sin ^2 x}{\cos x} = \frac{\cos^2x +\sin^2 x}{\cos x} $$ How do we get $\cos^2x$ in the numerator on the right-hand side?
I just don't get it.
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1Where is the cos^2 in the denominator? – Joao Dec 08 '14 at 01:33
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how about simplifying $2 + {3 \over 2}?$ – abel Dec 08 '14 at 01:51
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We need to find the common denominator, then add: $$\cos x + \frac {\sin^2 x}{\cos x} = \dfrac{\cos x}{\cos x}\cdot \frac {\cos x}{1} + \dfrac{\sin^2x}{\cos x}=\dfrac{\cos^2 x + \sin^2 x}{\cos x} = \dfrac 1{\cos x} = \sec x$$
amWhy
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this is kinda where I am lost though. common denom would be $$cosx$$, and we're not multiplying or dividing, so I'm confused how we get $$cos^2x$$ up top. – Jordan M Dec 07 '14 at 20:24
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I don't think that that is necessarily true:
Multiply through by cos(x)
$\cos^2x+ \sin ^2 x = \cos^3x + \sin^2x$
Using the identity $\cos^2x+ \sin ^2 x = 1$ we get:
$1 = \cos^3x + \sin^2x$
We can rearrange the identity to say $\sin ^2 x = 1 - \cos^2x $ and then substitute to get:
$1 = \cos^3x + (1 - \cos^2x)$
or:
$0 = \cos^3x - \cos^2x $
$\cos^3x = \cos^2x$
which only happens if cos(x) = 1 or cos(x) = 0 [the latter of which can't be true since cos(x) is in the denominator].
So it seems that the original expression is only true if cos(x) = 1.
Gabriel
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Whoops! Sorry. My computer loaded the problem as cos(x) + sin^2(x)/cos(x) = cos^2(x) + sin^2(x)/cos(x) – Gabriel Dec 07 '14 at 20:48
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ah then you're right, the person who edited my question edited it wrong. sorry about that! You were right if that was the question though, so ill give you an upvote :) – Jordan M Dec 07 '14 at 21:33