Question

help me please: cos(arcsin(3/5)+arctan(4)) So I have checked and doublechecked, tripple checked my work, and its still coming out wrong...

Answer 1

to evaluate cos(arcsin(3/5)+arctan(4)), we will use the identity

cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y)

in this case

x = arcsin(3/5)
y = arctan(4)

Answer 2

did you start out that way?

Answer 3

nope... i thought that once i found out what the cos was i can just add it

Answer 4

let me do it your way

Answer 5

give it a shot and post what you get

Answer 6

yay i got it right!!

Answer 7

i wasnt using the formula
thank u

Answer 8

so now that i have u can i ask u something else?

Answer 9

sure, go ahead

Answer 10

sin(pi/18) i have no idea how to solve this

Answer 11

it must remain in fraction form

Answer 12

ok one moment

Answer 13

Any ideas satellite? I'm trying to look for a formula dealing with sin(k*pi/6) where k = 1/3, but all I can find so far are instances where k is a whole number. I'll keep looking though.

Answer 14

yeah good luck with this program
i tried to solve
\[\frac{a}{6}+\frac{b}{3}=\frac{1}{18}\] but the problem with that is you are looking for integer solutions of
\[3a+6b=1\] and that is not going to work

Answer 15

i think you can do it by solving some sort of cubic equation but that is not going to be easy
in this case it is easier to work in degrees, and you are looking for \[\sin(10)\] a quick google search returns this result
https://answers.yahoo.com/question/index?qid=20080511045049AAOrNdH

Answer 16

seems cruel to have you solve a cubic

Answer 17

if you just type it in to wolfram you do not get the usual nested radicals etc, to make it seem like a simple application of a addition angle formula
instead you get it is the root of \( 8 x^3-6 x+1 \)

Answer 18

sure it is
\[\sin(\frac{\pi}{18})\]?

Answer 19

i did and i didnt understand how they resolved it.. sorr i meant sinpi/12

Answer 20

oooooooooooooooooooooooooooooooooooooh

Answer 21

sorry

Answer 22

i didnt notice i had typed it out incorrectly

Answer 23

\[\frac{\pi}{12}\] is half of \(\frac{\pi}{6}\) use the "half angle" formula

Answer 24

yeah i am trying to do that right now
but i am not sure how i would get cosA but let me see

Answer 25

you dont need to find cos(pi/12)

instead, find cos(pi/6)

Answer 26

all you need is the cosine of \(\frac{\pi}{6}\) which is not hard to find

Answer 27

ohhhh
i got it

Answer 28

since the formula is

\[\Large \sin\left(\frac{A}{2}\right) = \sqrt{\frac{1-\cos(A)}{2}}\]

some books state it as

\[\Large \sin(A) = \sqrt{\frac{1-\cos(2A)}{2}}\]

Answer 29

ok so simple basic question here.. i insert the numbers... and i get \[\frac{ \sqrt{1-(\sqrt{3}/2)} }{ 2 }\]

Answer 30

yes, now simplify

Answer 31

wouldnt there be a sq rt over the denominator?

Answer 32

i mean on the two

Answer 33

oh yeah it should be, just noticed that

Answer 34

but according to wolfram that is the answer

Answer 35

hmm let me see what I get

Answer 36

no sorry according to wolfram its sqrt2-sqrt3 / 2

Answer 37

one sec

Answer 38

ok this is what I get (it's a lot, but hopefully it all makes sense)

\[\Large \sin\left(\frac{\pi}{12}\right) = \sin\left(\frac{1}{2}*\frac{\pi}{6}\right)\]

\[\Large \sin\left(\frac{1}{2}*A\right) = \sqrt{\frac{1-\cos(A)}{2}}\]

\[\Large \sin\left(\frac{1}{2}*\frac{\pi}{6}\right) = \sqrt{\frac{1-\cos\left(\frac{\pi}{6}\right)}{2}}\]

\[\Large \sin\left(\frac{1}{2}*\frac{\pi}{6}\right) = \sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}\]

\[\Large \sin\left(\frac{1}{2}*\frac{\pi}{6}\right) = \sqrt{\frac{2}{2}*\frac{1-\frac{\sqrt{3}}{2}}{2}}\]

\[\Large \sin\left(\frac{1}{2}*\frac{\pi}{6}\right) = \sqrt{\frac{2*\left(1-\frac{\sqrt{3}}{2}\right)}{2*2}}\]

\[\Large \sin\left(\frac{1}{2}*\frac{\pi}{6}\right) = \sqrt{\frac{2*\left(1\right)-2*\left(\frac{\sqrt{3}}{2}\right)}{2*2}}\]

\[\Large \sin\left(\frac{1}{2}*\frac{\pi}{6}\right) = \sqrt{\frac{2-\sqrt{3}}{4}}\]

\[\Large \sin\left(\frac{1}{2}*\frac{\pi}{6}\right) = \frac{\sqrt{2-\sqrt{3}}}{\sqrt{4}}\]

\[\Large \sin\left(\frac{1}{2}*\frac{\pi}{6}\right) = \frac{\sqrt{2-\sqrt{3}}}{2}\]

\[\Large \sin\left(\frac{\pi}{12}\right) = \frac{\sqrt{2-\sqrt{3}}}{2}\]

Answer 39

wait why are u multiplying the two?

Answer 40

with this step here?

\[\Large \Large \sin\left(\frac{1}{2}*\frac{\pi}{6}\right) = \sqrt{\frac{2}{2}*\frac{1-\frac{\sqrt{3}}{2}}{2}}\]

Answer 41

yes

Answer 42

I multiplied top and bottom of the fraction by the LCD 2 to clear out that inner fraction

Answer 43

it's the same as multiplying by 1 since 2/2 = 1

Answer 44

ok got it makes more sense... thank u so much

Answer 45

yeah i know... i just didnt see it like that

Answer 46

glad its clicking now

Answer 47

thank u so so much

Answer 48

you're welcome