double angle formulas
cot \[5\pi \over 6\]
Do you want to solve it by using Double Angle Formulas ??
yes please
Can't you solve it directly ??
it says find the "exact" value
idk how to imply that
\[\frac{5 \pi}{6} = 150^{\circ}\] Right ??
you dont really need to use double angle formulas for this...
maybe if its 5pi/12 then yes it would advisable to use o.0
\[\cot(150) = \cot(90 + 60) = -\tan(60) = - \sqrt{3}\]
it's cot (5π)/6
\[\frac{\cot(5 \pi)}{6}\] 6 is not in the angle it is the fraction right???
yes that's right
how do you know it's −3√
is that your answer tell me or not ??
−√3
yes it is but i mean how do you come up with that answer ?
So your question is : \[\large \cot(\frac{5 \pi}{6})\] Right ??
yuppp i just didnt know how to type that in the equation format
See : You should convert it to degrees so that it becomes easy for you.. \[\frac{\pi}{6} = \frac{\pi}{6} \times \frac{180}{\pi} = 30^{\circ}\] So: \[\frac{5 \pi}{6} = 5 \times 30 = 150\] Getting till here ??
I always go back to the degree value of pi and the radian value of 360 degrees
So : you have not: \[\large \cot(150^{\circ}) \implies \cot(180^{\circ} - 30^{\circ}) \implies -\cot(30^{\circ})\] Can you find cot(30) ??
cot= -tan
\[\cot(30) = \tan(60) = \sqrt{3}\]
how do you know which values to use
What i have written above: -cot(30) so here you will use : cot(30) i guess..
k no you just confused me
Where ??
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