<s>[INST] Calculate $\sin^5\alpha-\cos^5\alpha$ if $\sin\alpha-\cos\alpha=\frac12$ Calculate $$\sin^5\alpha-\cos^5\alpha$$ if $\sin\alpha-\cos\alpha=\dfrac12$.
The main idea in problems like this is to write the expression that we need to calculate in terms of the given one (in this case we know $\sin\alpha-\cos\alpha=\frac12$).
I don't see how to even start to work on the given expression as we cannot use $a^2-b^2=(a-b)(a+b)$ or $a^3-b^3=(a-b)(a^2+ab+b^2)$. So in other words, I can't figure out how to factor the expression (even a little).

 The given answer is $\dfrac{79}{128}$.

 [/INST] Another approach: Solve $ \sin{z}-\cos{z}=\frac{1}{2}$
This results in $e^{iz}=\frac{\sqrt{7}-i}{2(i-1)}$
Plugging this into the complex sin and cos definition results in:
$\sin{z}=\frac{1-1\sqrt{7}}{4}$ and $\cos{z}= \frac{-1-1\sqrt{7}}{4}$
When applying this in $sin^5z-cos^5z$ all uneven powers of $\sqrt(7)$ cancel out.
What remains is: $sin^5z-cos^5z=\frac{1+70+5\cdot 49}{2^9}=\frac{79}{128}$
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