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4U Maths Question Thread

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shekhar.patel:
Hi. How do I do Q7     

fun_jirachi:
Hey there!

If I'm not mistaken, an assumption should be made that \(a, b \in \mathbb{Z}\). Notice that if a and b are integers, and they have an even difference, they must both be odd or both be even. Consider now what happens to the sum of a and b, and thus the difference of the squares.

Hope this helps :)

mrsc:
Hey guys, just needed help with this complex number question part (d). Thanks

fun_jirachi:
Hey there!

Consider what happens if you take \((\omega + \omega ^8)(\omega ^2 + \omega ^7)(\omega ^4+ \omega ^5)\), where \(\omega = \cos \frac{2\pi}{9} +i\sin \frac{2\pi}{9}\).

We have that \((\omega + \omega ^8)(\omega ^2 + \omega ^7)(\omega ^4+ \omega ^5) = (2\cos \frac{2\pi}{9})(2\cos \frac{4\pi}{9})(-2\cos \frac{\pi}{9})\).

We also have that by expanding the brackets and simplifying using part (b), \((\omega + \omega ^8)(\omega ^2 + \omega ^7)(\omega ^4+ \omega ^5) = -1\).

Hence, we have that \((2\cos \frac{2\pi}{9})(2\cos \frac{4\pi}{9})(-2\cos \frac{\pi}{9}) = -1 \implies \cos \frac{2\pi}{9} \cos \frac{4\pi}{9} \cos \frac{\pi}{9} = 1\).

What's evident in this question is that you occasionally need to look to manipulate the question into a nicer form - here, the solution only really became obvious for me after manipulating the result to \((2\cos \frac{2\pi}{9})(2\cos \frac{4\pi}{9})(-2\cos \frac{\pi}{9}) = -1\). From this sort of position, it's a lot simpler and much less time consuming to explore possible solutions, especially given a number of results from previous parts of the question. Try this line of thinking with similar questions!

Hope this helps :)

mrsc:
Thanks a lot for your help!!! I get it now. Also stuck with part (c) of another question. For these types of questions, I know how to solve the LHS but for the RHS I'm not sure. I tried changing the roots from the equation in part (b) into mod-arg form first and then dividing by z^3 but it didn't work.

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