Question

Use the properties of trig functions to find the exact value:
sin(-pi/12)csc(25pi/12)?

Answer 1

\[\sin{\frac{-\pi}{12}} \csc{\frac{25\pi}{12}} \\ -\sin{\frac{\pi}{12}} \csc{\frac{\pi}{12}} \\ -\sin{\frac{\pi}{12}} \frac{1}{\sin{\frac{\pi}{12}}} \\ -1 \]

Answer 2

note that \[\csc{\frac{25\pi}{12}} = \csc{(2\pi+\frac{\pi}{12})}= \csc{(\frac{\pi}{12})}\]

Answer 3

ahhh i see the csc\[25\pi \div12\] can have 24pi taken out so its just csc pi/12 and its the reciprocal of 1/sin pi/12, then cancel the two out and dont forget the negative sign, so its -1. got it. Thanks!

Answer 4

idk why i forgot about the periodic property!

Answer 5

>.> that happened to me when I do my exam..

Answer 6

yeah its tough memorizing all this in a 6 week accelerated precalc class 0_0 lol