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anonymous
 one year ago
find the exact value of cos75 degrees
anonymous
 one year ago
find the exact value of cos75 degrees

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Use a half angle identity (150/2 = 75) or sum and difference identity (30 + 45) = 75 half angle (Θ = 150°): \[\cos \frac{ \theta }{ 2 }=\sqrt{\frac{ 1+\cos \theta }{ 2 }}\] sum and difference (u = 30°, v = 45°) \[\cos (u + v) = \cos u \cos v  \sin u \sin v\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That is just as confusing as the notes

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is the example I have in the notes

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so it looks like they used a sum and difference identity: \(cos(u+v)= \cos u \cos v− \sin u \sin v\) The first thing you have to do is find 2 numbers on your unit circle that add up to 75. That's where the 30 and 45 come from. Then let u be 30 and let v be 45 and we're going to plug them in \(cos 75°=cos(30°+45°) With me so far?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have one as a tab, not printed out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So this is our identity cos (u + v) = cos u cos v  sin u sin v Plug in the 30 and 45 cos (30° + 45°) = cos 30° cos 45°  sin 30° sin 45° Now use the unit circle to find the 4 values in the identity: cos 30° cos 45° sin 30° sin 45°

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why do you subtract it by sin ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0your formula is different than the one in my notes :/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that's just the identity. Here's a list of the identities. Look under sum and difference http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so how do you know which one to use ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which sign to use, or which identity?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It depends on what you're trying to find. For this one you can use multiple identities. I started off wanting to use a half angle because then all you'd have to do is find cos 150° and plug it into the formula. You could also have use the difference identity for cosine because 120  45 = 75. You're just looking for any combination that works. They'll all give you the same answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so pretty much you can us any, just depending if you're adding or subtracting ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah as long as it's the correct trig function and the numbers add/subtract right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay , I understand that part so far

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cos (30° + 45°) = cos 30° cos 45°  sin 30° sin 45° ok so now we plug in the numbers from the unit circle \[\cos 75°=\frac{ \sqrt3 }{ 2 }\times \frac{ \sqrt2 }{ 2 }\frac{ 1 }{ 2 }\times \frac{ \sqrt2 }{ 2 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\cos 75°=\frac{ \sqrt6 }{ 4 }\frac{ \sqrt2 }{ 4 }\] \[\cos 75°=\frac{ \sqrt6 \sqrt2 }{ 4 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0where did the 1/2 go

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you're multiplying xD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't remember how to add and subtract radicals. I always mix it up

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the numbers under the radicals have to be the same to add or subtract, so this can't be simplified any further

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so it'd have to be \[\sqrt{4}  \sqrt{4} \] to subtract ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but wouldn't that just always equal 0 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but if you had something like this, just subtract the "coefficients" \[5\sqrt63\sqrt6=2\sqrt6\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oooh okay, it's coming back lol. It's been a longer summer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0haha everyone's saying that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I've been taking summer school for algebra 2 and trigonometry was thrown into algebra 2, which makes no sense to me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah that doesn't make sense. This is more advanced trig so it really shouldn't be mixed in there

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I passed both semester of algebra 2 with a D, that being an accelerated class . I'm taking regular algebra 2 for summer school . Idk how the state thinks this is possible for an algebra 2 class lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That sounds like GA. I've tutored here since 2009 and the curriculum has changed 4 times. They keep moving stuff from one class to the next. It's ridiculous

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Common core is just awful . This year is the first year for common core to be fully enforced by math . Trig is introduced in accelerated algebra 2 when I was taking the class , but it wasn't so in depth

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@peachpi could you help me with tan105 degrees ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\tan(uv)=\frac{ \tan u  \tan v }{ 1+\tan u \tan v }\] \[\tan(150°45°)=\frac{ \tan 150°  \tan 45° }{ 1+\tan 150° \tan 45° }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0to find tan , you have to divide sin by cos right ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and then you multiply the denominator out to only have a numerator ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay , ima try it and see what I come up with

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you divide by the reciprocal right ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so for tan150 , I got \[\frac{ 2 }{ 2\sqrt{3} }\] is that right ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0uh , no . where did that come from ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh, wait never mind. I misread what you had. That is correct for tan 150. I read tan 105

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh no lol. I'm doing it step by step

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so then tan45 is 1 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and to substitute , I got \[\frac{  \frac{ 2 }{ 2\sqrt{3} }  1 }{ 1+(\frac{ 2 }{ 2\sqrt{3} }) \times 1}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how do you factor that lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0start with simplifying: 2/(2√3) = 1/√3 Then multiply by √3/√3 to clear denominators dw:1440272933438:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oooh okay so you have to get rid of the denominator for tan150 fraction ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for this one yeah because you can't have the radical in the denominator

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0to rationalize the denominator , how do you do that ? ... xD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0jk , I looked at your answer and I figured it out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0multiply by the conjugate

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh ok. yeah foil will do it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so is the answer \[4\sqrt{6}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Somehow your numbers doubled. Should be \[2\sqrt3\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so when it's two radicals with the same number inside , it stays the same number ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440273681350:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440273726704:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh okay , I see what I did wrong
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