Help with trigonometry questions (1 Viewer)

Bob_man · Jan 18, 2022

hi im having some trouble figuring out how to find the solutions for questions 4, 8c and 8d.

Drongoski · Jan 18, 2022

Bob_man said:
hi im having some trouble figuring out how to find the solutions for questions 4, 8c and 8d. View attachment 34687 View attachment 34688

5uckerberg · Jan 19, 2022

Okay Since Drongoski has done Q8 for you I will take over for Q4. With Q4 You see here that
$2\cos{\left(2x-\frac{\pi}{3}\right)}=\sqrt{3}$
$\cos{\left(2x-\frac{\pi}{3}\right)}=\frac{\sqrt{3}}{2}$
For $0 \leq{x}\leq{\pi}$
We can turn this into $-\frac{\pi}{3}\leq{2x-\frac{\pi}{3}\leq{\frac{5\pi}{3}$
$\left(2x-\frac{\pi}{3}\right)=\frac{\pi}{6}, -\frac{\pi}{6}$
$2x=\frac{\pi}{2}, \frac{\pi}{6}$
$x=\frac{\pi}{4}, \frac{\pi}{12}$
I can see that D is correct but I just can't resist the fact that Fitzpatrick and his writers did not spot a typo where instead of 145 degrees it should be just 45 degrees and 15 degrees.

Note this if you are using the New Senior mathematics textbook keep an eye out on the typos there will be some there, another case of you cannot fully trust the textbook. A technique both students and teachers should learn when they use textbooks.

Bob_man · Jan 20, 2022

Thanks heaps!!

Help with trigonometry questions (1 Viewer)

Bob_man

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Drongoski

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Bob_man

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