General Identities (in Trigonometry) ADDITION AND SUBTRACTION IDENTITIES FIND THE VALUE OF: cos π/12
cos(π/12)=cos(π/4 - π/6) and then you can use compound angle identities
or use 1/12= 1/3-1/4
the difference formula for cosine \[\cos(\theta-\phi)=\cos\theta\cos\phi+\sin\theta\sin\phi\] (i still always have to look this up)
is cos π/12 = cos 15 degrees?
yes
yes, but that doesn't use the difference(subtraction) identity if you use the difference formula then you will get the exact result (with square roots and such)
oh i see how is this done then? i dont get what you're saying above with all those fractions
cos(A - B) = cosAcosB + sinBsinA cos(π/4 - π/6)=(cosπ/4)(cosπ/6) + (sinπ/4)(sinπ/6) and then you can use special triangles or whichever way you were taught
The trick in this problem is to recognise the fraction 1/12 can be expressed as a difference of simpler fractions, (use either damoss, or my conversion) then apply the difference formula, the trig identities of the simpler fraction are 'known', or can be found eaisly
why is it 1/3-1/4? i dont get that? @UnkleRhaukus
@damoss how did you get cos(π/4 - π/6)?
hm okay it may be easier for you if you're more comfortable with degrees to convert it to degrees. So π=180, yes? Then π/4=180/4=45 degrees and π/6=180/6=30 ; then 45 - 30=15
1/3-1/4 = 4/12 - 3/12 = 1/12
1/4 - 1/6 = 6/24 - 4/24 = 2/24 = 1/12
cos15=cos(45 - 30) if you will
(yeah it is a bit easier in degrees)
i see
will the answer be: √(6)/4 - √(2)/4 ? @UnkleRhaukus @damoss
i believe that negative should be a positive
make sure you used the correct identity, cos(A - B) = cosAcosB + sinBsinA
√(6)/4 + √(2)/4 is that it already? @damoss
yep
okay thanks :)
That is right, good work but you can simplify a little bit further √(2)/4 is a common factor
√(2)/4 (√(3) + 1 ) so is this the final answer? @UnkleRhaukus
YES top stuff!
okay thanks @UnkleRhaukus
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