Trig problems

This is about addition formulae

Solve the following for 0 <= x <= 2pi

sin x = 3 cos (x - pi/6)

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This is what I did (Wrong!)

sin x = 3 (cos x cos pi/6 + sin x sin pi/6)

sin x = 3 cos x cos pi/6 + 3 sin x sin pi/6

sin x - 3 sin x sin pi/6 = 3 cos x cos pi/6

sin x (1 - 3 sin pi/6) = 3 cos x cos pi/6

1 - 3 sin pi/6 = 3 cot x cos pi/6

and now I'm stuck...

help..

Jorge

Reply 1

m:)ckel

14

Well, sin(pi/6) = 0.5, and cos(pi/6) = (rt3)/2

So, that should help. From then on, it's just a matter of rearranging for cotx, and then solving.

Reply 2

El Stevo

11

sinx = 3cos(x-pi/6)

cos(x-pi/6) = cos(x)cos(pi/6) - sin(x)sin(pi/6)
cos(x-pi/6) = cos(x)sqrt3/2 - sin(x)/2

sinx = 3[cos(x)sqrt3/2 - sin(x)/2]
sinx = {(3sqrt3)/2}cosx - (3/2)sinx
5/2sinx = {(3sqrt3)/2}cosx
tanx = {(3sqrt3)/2}/(5/2)
tanx = {(3sqrt3)/2}.(2/5)
tanx = {(6sqrt3)/10}
tanx = {(3sqrt3)/5

x = 0.805
x = 3.95

Reply 3

habosh

13

El Stevo

sinx = 3cos(x-pi/6)

cos(x-pi/6) = cos(x)cos(pi/6) - sin(x)sin(pi/6)
cos(x-pi/6) = cos(x)sqrt3/2 - sin(x)/2

sinx = 3[cos(x)sqrt3/2 - sin(x)/2]
sinx = {(3sqrt3)/2}cosx - (3/2)sinx
5/2sinx = {(3sqrt3)/2}cosx
tanx = {(3sqrt3)/2}/(5/2)
tanx = {(3sqrt3)/2}.(2/5)
tanx = {(6sqrt3)/10}
tanx = {(3sqrt3)/5

x = 0.805
x = 3.95

cosx cos(pi/6)+sinxsin(pi/6)

Reply 4

Jorge

OP

8

El Stevo

sinx = 3cos(x-pi/6)

cos(x-pi/6) = cos(x)cos(pi/6) - sin(x)sin(pi/6)
cos(x-pi/6) = cos(x)sqrt3/2 - sin(x)/2

sinx = 3[cos(x)sqrt3/2 - sin(x)/2]
sinx = {(3sqrt3)/2}cosx - (3/2)sinx
5/2sinx = {(3sqrt3)/2}cosx
tanx = {(3sqrt3)/2}/(5/2)
tanx = {(3sqrt3)/2}.(2/5)
tanx = {(6sqrt3)/10}
tanx = {(3sqrt3)/5

x = 0.805
x = 3.95

Thanks everybody, I got it

sinx = 3(cosx sqrt3/2 + sinx/2)

-sinx/2 = cosx 3sqrt3/2

sinx = - cos 3sqrt3

tanx = - 3sqrt3 = -1.38

between 0 and 2pi > 1.76 and 4.90

Thanks

Reply 5

habosh

13

Jorge

Thanks everybody, I got it

sinx = 3(cosx sqrt3/2 + sinx/2)

-sinx/2 = cosx 3sqrt3/2

sinx = - cos 3sqrt3

tanx = - 3sqrt3 = -1.38

between 0 and 2pi > 1.76 and 4.90

Thanks

yep same as my answer :wink:

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