Question

find the exact value of cos75 degrees

Answer 1

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°)
\[\cos (u + v) = \cos u \cos v - \sin u \sin v\]

Answer 2

That is just as confusing as the notes

Answer 3

this is the example I have in the notes

Answer 4

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?

Answer 5

so far, yes lol

Answer 6

I have one as a tab, not printed out

Answer 7

that's good enough.

Answer 8

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°

Answer 9

why do you subtract it by sin ?

Answer 10

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

Answer 11

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

Answer 12

so how do you know which one to use ?

Answer 13

which sign to use, or which identity?

Answer 14

identity

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

okay , I understand that part so far

Answer 19

cos (30° + 45°) = cos 30° cos 45° - sin 30° sin 45°

ok so now we plug in the numbers from the unit circle
\[\cos 75°=\frac{ \sqrt3 }{ 2 }\times \frac{ \sqrt2 }{ 2 }-\frac{ 1 }{ 2 }\times \frac{ \sqrt2 }{ 2 }\]

Answer 20

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

Answer 21

where did the 1/2 go

Answer 22

jk

Answer 23

you're multiplying xD

Answer 24

yeah :)

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

right

Answer 29

but wouldn't that just always equal 0 ?

Answer 30

yes

Answer 31

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

Answer 32

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

Answer 33

haha everyone's saying that

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

@peachpi could you help me with tan105 degrees ?

Answer 40

105 = 150 - 45

Answer 41

\[\tan(u-v)=\frac{ \tan u - \tan v }{ 1+\tan u \tan v }\]
\[\tan(150°-45°)=\frac{ \tan 150° - \tan 45° }{ 1+\tan 150° \tan 45° }\]

Answer 42

yeah?

Answer 43

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

Answer 44

yes

Answer 45

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

Answer 46

right

Answer 47

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

Answer 48

ok

Answer 49

you divide by the reciprocal right ?

Answer 50

multiply *

Answer 51

yes

Answer 52

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

Answer 53

uh , no . where did that come from ?

Answer 54

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

Answer 55

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

Answer 56

so then tan45 is 1 ?

Answer 57

yes

Answer 58

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

Answer 59

Yes that's right

Answer 60

how do you factor that lol

Answer 61

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