# What am I misunderstanding with this simple Trigonometry question?

Simplify: $\sin^4x + \sin^2x \cdot \cos^2x$

The textbook states the answer as $\sin^2x$ and I understand the reasoning: Take a factor of $\sin^2x$ out and you are left with $\sin^2x \cdot 1$

However I can't work out why my method is wrong (it produces the answer of 1):

Divide everything by $\sin^2x$

$(\sin^4x / \sin^2x)+ (\sin^2x \cdot \cos^2x )/ \sin^2x$

Which cancels down to:

$\sin^2x + \cos^2x$

Which equals $1$.

I managed to get both results when using WolframAlpha to check my working! What am I misunderstanding?

Thanks!

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You cannot divide by $\sin^2 x$ without also multiply by $\sin^2 x$... It's not an equation. – Pacciu Mar 26 '11 at 18:27

You did nothing wrong. Except misunderstand what you did.

Let us say the simplification is some variable called $k(x)$. Then,

$$\sin^4 x +\sin^2 x \cos^2 x = k(x)$$

divide by $\sin^2 x$, so,

$$1 = \frac{k(x)}{\sin^2 x}$$

and thus,

$$k(x)=\sin^2 x$$

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Ah thank you, that's very clearly explained! – Danny King Mar 26 '11 at 18:31
well, except that this method is valid only when $\sin^2x\neq0$ – Andrea Mori Mar 26 '11 at 18:49
@ Andrea: the point is to let Danny realize what he did wrong. Not to tell him when the method is valid, the other solution is the best approach. – picakhu Mar 26 '11 at 18:55
Thanks to you both. Andrea made me realize I shouldn't fall back on my method in exams! – Danny King Mar 26 '11 at 19:31