Question

prove the following trig identity 2/(sqrt 3 cos x +sin x)= sec((pi/6)-x) help please!!!

Answer 1

......please

Answer 2

Is it ?

\[\frac{2}{\sqrt{3} \cos(x) + \sin(x)}\]

Answer 3

It should be LHS..

Answer 4

Dividing numerator and denominator by 2 will start the solution..

Answer 5

Yes that is the correct equation and the way I learned how to do it is that you have to make the left hand side equal the right. I will try dividing by 2.

Answer 6

I still do not understand it.

Answer 7

ok, can we go from the RHS? it's easier

Answer 8

\[\sec (\frac{ \pi }{ 6 }- x ) = \frac{ 1 }{ \cos (\frac{ \pi }{ 6 } -x)}\]

Answer 9

= \[\frac{ 1 }{ \cos \frac{ \pi }{ 6 } \cos x + \sin \frac{ \pi }{ 6 } \sin x}\]

Answer 10

= \[\frac{ 1 }{ \sqrt{3} /2\cos x + \frac{ 1 }{ 2 }\sin x}\]

Answer 11

take 1/2 up, you got LHS

Answer 12

Ok where did sin pi/6 come from? Wouldn't it be 1/(cos(pi/6)-cos x)?

Answer 13

nope, it's the formula cos (a- b) = cos a cos b+ sin a sin b
your a is pi/6 your b is x

Answer 14

kapeech???

Answer 15

So with the half in the denominator, do you multiply the numerator and denominator by 2?

Answer 16

you can do that. the same answer

Answer 17

Oh ok thank you so much!! I get it now.

Answer 18

yw