anonymous
  • anonymous
find the exact value of cos75 degrees
Mathematics
schrodinger
  • schrodinger
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
Use 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
  • anonymous
That is just as confusing as the notes
anonymous
  • anonymous
this is the example I have in the notes

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
ok 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
  • anonymous
so far, yes lol
anonymous
  • anonymous
I have one as a tab, not printed out
anonymous
  • anonymous
that's good enough.
anonymous
  • anonymous
So 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
  • anonymous
why do you subtract it by sin ?
anonymous
  • anonymous
your formula is different than the one in my notes :/
anonymous
  • anonymous
that'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
  • anonymous
so how do you know which one to use ?
anonymous
  • anonymous
which sign to use, or which identity?
anonymous
  • anonymous
identity
anonymous
  • anonymous
It 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
  • anonymous
so pretty much you can us any, just depending if you're adding or subtracting ?
anonymous
  • anonymous
yeah as long as it's the correct trig function and the numbers add/subtract right
anonymous
  • anonymous
okay , I understand that part so far
anonymous
  • anonymous
cos (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
  • anonymous
\[\cos 75°=\frac{ \sqrt6 }{ 4 }-\frac{ \sqrt2 }{ 4 }\] \[\cos 75°=\frac{ \sqrt6 -\sqrt2 }{ 4 }\]
anonymous
  • anonymous
where did the 1/2 go
anonymous
  • anonymous
jk
anonymous
  • anonymous
you're multiplying xD
anonymous
  • anonymous
yeah :)
anonymous
  • anonymous
I don't remember how to add and subtract radicals. I always mix it up
anonymous
  • anonymous
the numbers under the radicals have to be the same to add or subtract, so this can't be simplified any further
anonymous
  • anonymous
so it'd have to be \[\sqrt{4} - \sqrt{4} \] to subtract ?
anonymous
  • anonymous
right
anonymous
  • anonymous
but wouldn't that just always equal 0 ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
but if you had something like this, just subtract the "coefficients" \[5\sqrt6-3\sqrt6=2\sqrt6\]
anonymous
  • anonymous
oooh okay, it's coming back lol. It's been a longer summer
anonymous
  • anonymous
haha everyone's saying that
anonymous
  • anonymous
I've been taking summer school for algebra 2 and trigonometry was thrown into algebra 2, which makes no sense to me
anonymous
  • anonymous
yeah that doesn't make sense. This is more advanced trig so it really shouldn't be mixed in there
anonymous
  • anonymous
I 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
  • anonymous
That 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
  • anonymous
Common 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
  • anonymous
@peachpi could you help me with tan105 degrees ?
anonymous
  • anonymous
105 = 150 - 45
anonymous
  • anonymous
\[\tan(u-v)=\frac{ \tan u - \tan v }{ 1+\tan u \tan v }\] \[\tan(150°-45°)=\frac{ \tan 150° - \tan 45° }{ 1+\tan 150° \tan 45° }\]
anonymous
  • anonymous
yeah?
anonymous
  • anonymous
to find tan , you have to divide sin by cos right ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
and then you multiply the denominator out to only have a numerator ?
anonymous
  • anonymous
right
anonymous
  • anonymous
okay , ima try it and see what I come up with
anonymous
  • anonymous
ok
anonymous
  • anonymous
you divide by the reciprocal right ?
anonymous
  • anonymous
multiply *
anonymous
  • anonymous
yes
anonymous
  • anonymous
so for tan150 , I got \[-\frac{ 2 }{ 2\sqrt{3} }\] is that right ?
anonymous
  • anonymous
uh , no . where did that come from ?
anonymous
  • anonymous
oh, wait never mind. I misread what you had. That is correct for tan 150. I read tan 105
anonymous
  • anonymous
oh no lol. I'm doing it step by step
anonymous
  • anonymous
so then tan45 is 1 ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
and to substitute , I got \[\frac{ - \frac{ 2 }{ 2\sqrt{3} } - 1 }{ 1+(-\frac{ 2 }{ 2\sqrt{3} }) \times 1}\]
anonymous
  • anonymous
Yes that's right
anonymous
  • anonymous
how do you factor that lol
anonymous
  • anonymous
start with simplifying: -2/(2√3) = -1/√3 Then multiply by √3/√3 to clear denominators |dw:1440272933438:dw|
anonymous
  • anonymous
oooh okay so you have to get rid of the denominator for tan150 fraction ?
anonymous
  • anonymous
for this one yeah because you can't have the radical in the denominator
anonymous
  • anonymous
thank you!
anonymous
  • anonymous
you're welcome
anonymous
  • anonymous
to rationalize the denominator , how do you do that ? ... xD
anonymous
  • anonymous
jk , I looked at your answer and I figured it out
anonymous
  • anonymous
multiply by the conjugate
anonymous
  • anonymous
oh ok. yeah foil will do it
anonymous
  • anonymous
so is the answer \[-4-\sqrt{6}\]
anonymous
  • anonymous
Somehow your numbers doubled. Should be \[-2-\sqrt3\]
anonymous
  • anonymous
so when it's two radicals with the same number inside , it stays the same number ?
anonymous
  • anonymous
|dw:1440273681350:dw|
anonymous
  • anonymous
|dw:1440273726704:dw|
anonymous
  • anonymous
oh okay , I see what I did wrong
anonymous
  • anonymous
thank you again
anonymous
  • anonymous
you're welcome

Looking for something else?

Not the answer you are looking for? Search for more explanations.