Question

cos(x+y)=0.17 and cosx=0.50, find siny

Answer 1

Use the angle sum formula for cosine
\[\cos(x+y) = \cos(x) \cos(y) - \sin(x) \sin(y)\]
Use pythagorean identity to find sin(x),cos(y)
\[\sin^{2}(x) = 1-\cos^{2} (x)\]
\[\cos^{2} (y) = 1-\sin^{2}(y)\]

Answer 2

\[0.17=\frac{ cosy - \sqrt{3}siny }{ 2 }\]

that is what i got after using sum formula for cosine.. i don't know what to do next

Answer 3

you need to replace cos(y) with an expression in terms of sin(y) since that is what you are trying to find

look at pythagorean identity

Answer 4

\[cosy=\sqrt{1-\sin ^{2}y}\]

is that it?? i will substitute that?

Answer 5

yep

Answer 6

\[0.34=\sqrt{1-\sin ^{2}y}-\sqrt{3}siny\]

that's what i got.. and i don't know what to do next. will i square both sides of the equation?

Answer 7

you got it...i recommend moving the " sqrt3 siny" term to other side first

Answer 8

after you square both sides...combine like terms and set equal to zero

then you will need to use quadratic formula

Answer 9

ok.. i'll try to do that.. i'll post my answer afterwards and check it if its right.. thanks!! :))

Answer 10

ok yw

Answer 11

@dumbcow please help me!! :) http://openstudy.com/study#/updates/505dd105e4b02e1394100030

Answer 12

i got a problem with my calculator :(( can you give me some site where i can best answer quadratic formula?? :((

Answer 13

wolframalpha.com will work as calculator

http://www.wolframalpha.com/input/?i=4u^2+%2B.68sqrt%283%29u+%2B%28.34^2+-1%29+%3D+0

i replaced sin(y) with u

Answer 14

its simple cosx=0.5 that eans 1/2 hence x=60

now cos(x+y)=0.17

cos(60+y)=0.17

60+y=arccos(0.17)

y=arccos(0.17)-60

cosy=cos(arccos(0.17)-60)

Answer 15

sriramkumar did it easier :O
i got the answer right away..
thanks to you two.. :))