Question

prove the following trigonometric identities. 2/(rt(3)cos(x)+sin(x))=sec(pi/6-x)
tan(x/2)=sin(x)/(1+cos(x))

Answer 1

@dinamix

Answer 2

\[\frac{ 2 }{\sqrt{3} \cos(x)+\sin(x)}=\sec(\pi/6-x) \tan(x/2)=\frac{ \sin (x) }{1+\cos (x) } \]
is that your question?

Answer 3

no they are two separate equations.
2/(rt(3)cos(x)+sin(x))=sec(pi/6-x)
tan(x/2)=sin(x)/(1+cos(x))

Answer 4

oK

Answer 5

I will do only one, which one do you like?

Answer 6

do the top one

Answer 7

you can do either one if you want

Answer 8

for the second
\[tan\frac{x}{2}=\frac{sinx}{1+cosx}\]
try this
\[tan\frac{x}{2}=\frac{2sin\frac{x}{2}cos \frac{x}{2}}{1+(1-2sin^2 \frac{x}{2})}\]

Answer 9

sec (pi/6 - x)

= 1 / (sqrt3/2 cos x + sin pi/6 sin x)

= 2 / (sqrt 3 cos + sin x)

Answer 10

* I missed out the second step which is

= 1 / (cos pi/6 cos x + sin pi/6 sin x)

Answer 11

\[\frac{ 2 }{\sqrt{3} \cos(x)+\sin(x)}=\sec(\pi/6-x)\]
\[\frac{ 2 }{\sqrt{3} \cos(x)+\sin(x)}=\frac{ 1 }{ \cos{(\pi/6-x)} }\]
using:cos(u-v)=cos u cos v+sin u sin v
\[\frac{ 2 }{\sqrt{3} \cos(x)+\sin(x)}=\frac{ 1 }{ \cos{(\pi/6) \times \cos(x)}+\sin{(\pi/6) \times \sin(x)} }\]

Answer 12

i got that far ASAAD123 put i didn't know what to do after that

Answer 13

the denominator = cos (pi/6 - x)

Answer 14

and 1 / cos = sec

Answer 15

wouldn't you also change sin(pi/6) to 1/2

Answer 16

no

its the sin of a compound angle (pi/6 - x) not sin pi/6

Answer 17

ok

Answer 18

so how do you get the sides to equal

Answer 19

i had to use sin pi/6 = 1/2 because i started with RHS and converted to LHS whereas ASAAD did the reverse

Answer 20

with these identities you choose one side and try to convert it to the other. This proves the identity.

Answer 21

ok

Answer 22

i think i understand it now

Answer 23

so you can use asaad's or mine. Either would do.

Answer 24

There is only more more step to prove IrishBoy's solution

Answer 25

i already got irish boy's solution