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There are two terms I have trouble with simplifying:

1.)

$\dfrac{\tan^2(\gamma) -\cot^2(\gamma)}{4(2\sin^2(\gamma) - 1)} = ?$

The result is supposed to be $\dfrac{1}{4}$.

2.)

$2\sin^2(\delta) + \cos^4(\delta) - \sin^4(\delta) = ?$

For this one the result is supposed to be $1$.

How do I simplify these terms?

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For the second: use $x^4 - y^4 = (x^2 - y^2)(x^2 + y^2)$ –  TMM May 20 '12 at 15:55
    
There are easier ways of doing this, but since everything is squared there is a very obvious identity to try. You could use this to write everything in terms of $\sin^2 (\delta)$ - which is a technique you can use if nothing else comes to mind. –  Mark Bennet May 20 '12 at 15:58
    
I don't think you can simplify the first one. Are you sure you got it right? –  Milosz Wielondek May 20 '12 at 16:09
    
@Milosz: Yeah, the textbook says so. The result of the first term should be $\dfrac{1}{4}$, I just don't know why. –  Miroslav Cetojevic May 20 '12 at 16:33
    
@MiroslavCetojevic it's definitely wrong, try substituting any value for $\gamma$ and you'll see it doesn't hold. See my edit to my answer. –  Milosz Wielondek May 20 '12 at 16:39
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1 Answer

up vote 2 down vote accepted

For the first one, I don't think it can be simplified more than to $\frac{1}{4}\sec\gamma\csc\gamma=\csc^2(2\gamma)$, and it certainly won't simplify to $\frac{1}{4}$ (try substituting any value for $\gamma$, and you'll see that the result varies with $\gamma$, hence it can't simplify to a constant).

For the second one write it like TMM suggested and use the pythagorean identity $\cos^2{\delta}+\sin^2\delta=1$:

$$ \begin{align} 2\sin^2\delta + \cos^4\delta - \sin^4\delta =& 2\sin^2\delta+(\cos^2\delta-\sin^2\delta)(\cos^2\delta+\sin^2\delta)\\ =&\cos^2\delta+\sin^2\delta\\ =&1 \end{align} $$

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There is also $2\sin^2 \delta + \cos^4 \delta - \sin^4\delta = 2\sin^2\delta+(1-\sin^2\delta)^2 - \sin^4 \delta = 1$ –  Mark Bennet May 20 '12 at 16:46
    
Is writing $sec$, $csc$ rather than $\sin$, $\cos$ intentional or a typo? –  Miroslav Cetojevic May 20 '12 at 17:42
    
@MiroslavCetojevic Intentional. Note that $\sec\gamma=\frac{1}{\cos\gamma}$ and $\csc\gamma=\frac{1}{\sin\gamma}$. –  Milosz Wielondek May 20 '12 at 18:10
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