Question

Prove trig identity: 2 / (sqrt3 cos x + sin x) = sec((pi/6)-x)

Answer 1

write sec in cos
then use cos(a-b)=cosacosb+sinasinb
simplify
and u will get the solution

guaranteed

Answer 2

@ikram002p hey, how are you!? Could you help on my question?

Answer 3

im fine :) , wbu lol :P

Answer 4

I'm good lol haven't been on here in awhile

Answer 5

yeah me too

Answer 6

:) do you think you could help? I just have a few questions. I really appreciated your help last time

Answer 7

im kid off had headach lol , cant think :O
sry my prain is freezing :)
good luck

Answer 8

I really need the help! Do you know anybody that could?

Answer 9

@Mashy hey could you help me filling in the table?

Answer 10

\[\sqrt{3 cosx + \sin x} \ \ \ \ ?\]

Answer 11

no just sqrt3

Answer 12

oh wait \[\frac{2}{\sqrt{3cosx + sinx}}\]

Answer 13

plz write using equation editor :P

Answer 14

Ok hold on :)

Answer 15

\[2\div (\sqrt{3}\cos x +\sin x) = \sec((\Pi/6)-x)\]

Answer 16

oh.. ok ok

there is a root 3..

that should give you a hint ;-)

Answer 17

I really need to see the steps lol I'm not really good at this :)

Answer 18

I've been trying to get help for an hour lol

Answer 19

do you know the identity

cos(A)cos(B) + sin(A)sin(B) = ?

Answer 20

= cos(A-B)

Answer 21

yea.. so
now.. look at the denominator..
can u convert it into that form?

Answer 22

probably but I couldn't tell you how. I'm going to school in 30 min so I would really appreciate an answer as long as you explain why so I can understand. Should I solve for calculations before reasons?

Answer 23

for me, for the bcos(x) + asin(x) can be converted into k cos(alpha - x) form.
with k = sqrt(a^2 + b^2) and alpha = arctan(a/b)

Answer 24

I have to fill in the table at the top. I don't know how to convert all of that

Answer 25

i dont know how to do with that table, but you can find the relation of it with my steps :) see the equation of (sqrt3 cos x + sin x), here a = 1 and b = sqrt(3).
so, k = sqrt(a^2+b^2) = sqrt(1^2 + (sqrt(3))^2) = sqrt(4) = 2
and to get alpha, use
tan(alpha) =arctan(a/b) = arctan(1/sqrt(3)) = 30 degrees = pi/6
(note : a and b are positive then alpha in the first quadrant)

back to equation above, see the left side :
2 / (sqrt3 cos x + sin x) = 2 / 2cos(pi/6 - x) = 1/cos(pi/6 - x) = sec(pi/6 - x)

QED

Answer 26

Thanks for this, but I don't think it helps because it doesn't fit in the table. WOuld this help you: http://openstudy.com/study#/updates/516cadb4e4b02ec89c5b336f

Answer 27

i saw it, but they didnt use the table too :)
i just prove from the left side, while they start from the right side.

Answer 28

So would my three calculations be:1. sec(pi/6 - x) = 1/cos(pi/6-x)

2. = 1 / cos pi/6 cosx + sin pi/6 sinx
3. = 1 / (sqrt3 / 2cos x) + 1/2 sinx

Answer 29

That's the answer to the calculation side?

Answer 30

maybe, then just simplify by multiply by 2
4. => 2/sqrt(3)cosx+sinx
but your table only incluads 3 rows, lol

Answer 31

is there a step I could take away to add your 4th step?