Use the sum or different to find he exact value of each trigonometric function a) tan(π)/(12)
tan(pi/12)=tan(pi/6 - pi/12)
a certain sense of deja vu
what answer?
tan pi/2=15 dgree
tan15=tan(45-30)
forget degrees, this has nothing to do with degrees
i started to answer in last post, but only got half way, when mandolino linked to previous post here http://openstudy.com/users/mandolino#/users/mandolino/updates/4e86d5720b8b7ce881cf43a3 where it is worked out in all details
\[\tan 15=\frac{\tan45-\tan30}{1-\tan45\tan30}\]
"final answer" is \[2-\sqrt{3}\] but he or she wrote all the steps
from here I have a trouble to solve
\[2-\sqrt{3}\]is correct I don't know how get that answer
you are confusing yourself by thinking you have to switch to degrees. but you have the formula backwards
hold on let me find the right post
thank you for the link I can find the link I post early
\[\tan(\frac{\pi}{3}- \frac{\pi}{4})=\frac{\tan(\frac{\pi}{3})-\tan(\frac{\pi}{4})}{1-\tan(\frac{\pi}{3})\tan(\frac{\pi}{4})}\]
so you need the following numbers to put in the formula \[\tan(\frac{\pi}{3})=\sqrt{3}\] and \[\tan(\frac{\pi}{4})=1\]
put those numbers directly into the "subtraction angle" formula to get \[\tan(\frac{\pi}{12})=\frac{\sqrt{3}-1}{1-\sqrt{3}}\] and then rationalize the denominator to get your answer
ty
yw, hope it is clear
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