Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
1  
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
add comment

1 Answer

up vote 6 down vote accepted

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,

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

share|improve this answer
    
Ah thank you, that's very clearly explained! –  Danny King Mar 26 '11 at 18:31
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.