find the exact value of cos75 degrees

- anonymous

find the exact value of cos75 degrees

- schrodinger

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- 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

That is just as confusing as the notes

- anonymous

this is the example I have in the notes

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## More answers

- 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

so far, yes lol

- anonymous

I have one as a tab, not printed out

- anonymous

that's good enough.

- 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

why do you subtract it by sin ?

- anonymous

your formula is different than the one in my notes :/

- 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

so how do you know which one to use ?

- anonymous

which sign to use, or which identity?

- anonymous

identity

- 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

so pretty much you can us any, just depending if you're adding or subtracting ?

- anonymous

yeah as long as it's the correct trig function and the numbers add/subtract right

- anonymous

okay , I understand that part so far

- 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

\[\cos 75°=\frac{ \sqrt6 }{ 4 }-\frac{ \sqrt2 }{ 4 }\]
\[\cos 75°=\frac{ \sqrt6 -\sqrt2 }{ 4 }\]

- anonymous

where did the 1/2 go

- anonymous

jk

- anonymous

you're multiplying xD

- anonymous

yeah :)

- anonymous

I don't remember how to add and subtract radicals. I always mix it up

- anonymous

the numbers under the radicals have to be the same to add or subtract, so this can't be simplified any further

- anonymous

so it'd have to be \[\sqrt{4} - \sqrt{4} \] to subtract ?

- anonymous

right

- anonymous

but wouldn't that just always equal 0 ?

- anonymous

yes

- anonymous

but if you had something like this, just subtract the "coefficients"
\[5\sqrt6-3\sqrt6=2\sqrt6\]

- anonymous

oooh okay, it's coming back lol. It's been a longer summer

- anonymous

haha everyone's saying that

- anonymous

I've been taking summer school for algebra 2 and trigonometry was thrown into algebra 2, which makes no sense to me

- anonymous

yeah that doesn't make sense. This is more advanced trig so it really shouldn't be mixed in there

- 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

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

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

@peachpi could you help me with tan105 degrees ?

- anonymous

105 = 150 - 45

- 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

yeah?

- anonymous

to find tan , you have to divide sin by cos right ?

- anonymous

yes

- anonymous

and then you multiply the denominator out to only have a numerator ?

- anonymous

right

- anonymous

okay , ima try it and see what I come up with

- anonymous

ok

- anonymous

you divide by the reciprocal right ?

- anonymous

multiply *

- anonymous

yes

- anonymous

so for tan150 , I got \[-\frac{ 2 }{ 2\sqrt{3} }\] is that right ?

- anonymous

uh , no . where did that come from ?

- anonymous

oh, wait never mind. I misread what you had. That is correct for tan 150.
I read tan 105

- anonymous

oh no lol. I'm doing it step by step

- anonymous

so then tan45 is 1 ?

- anonymous

yes

- anonymous

and to substitute , I got \[\frac{ - \frac{ 2 }{ 2\sqrt{3} } - 1 }{ 1+(-\frac{ 2 }{ 2\sqrt{3} }) \times 1}\]

- anonymous

Yes that's right

- anonymous

how do you factor that lol

- anonymous

start with simplifying: -2/(2√3) = -1/√3
Then multiply by √3/√3 to clear denominators
|dw:1440272933438:dw|

- anonymous

oooh okay so you have to get rid of the denominator for tan150 fraction ?

- anonymous

for this one yeah because you can't have the radical in the denominator

- anonymous

thank you!

- anonymous

you're welcome

- anonymous

to rationalize the denominator , how do you do that ? ... xD

- anonymous

jk , I looked at your answer and I figured it out

- anonymous

multiply by the conjugate

- anonymous

oh ok. yeah foil will do it

- anonymous

so is the answer \[-4-\sqrt{6}\]

- anonymous

Somehow your numbers doubled. Should be
\[-2-\sqrt3\]

- anonymous

so when it's two radicals with the same number inside , it stays the same number ?

- anonymous

|dw:1440273681350:dw|

- anonymous

|dw:1440273726704:dw|

- anonymous

oh okay , I see what I did wrong

- anonymous

thank you again

- anonymous

you're welcome

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