anonymous
  • anonymous
prove the following trigonometric identities. 2/(rt(3)cos(x)+sin(x))=sec(pi/6-x) tan(x/2)=sin(x)/(1+cos(x))
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@dinamix
anonymous
  • anonymous
\[\frac{ 2 }{\sqrt{3} \cos(x)+\sin(x)}=\sec(\pi/6-x) \tan(x/2)=\frac{ \sin (x) }{1+\cos (x) } \] is that your question?
anonymous
  • anonymous
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))

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
oK
anonymous
  • anonymous
I will do only one, which one do you like?
anonymous
  • anonymous
do the top one
anonymous
  • anonymous
you can do either one if you want
IrishBoy123
  • IrishBoy123
for the second \[tan\frac{x}{2}=\frac{sinx}{1+cosx}\] try this \[tan\frac{x}{2}=\frac{2sin\frac{x}{2}cos \frac{x}{2}}{1+(1-2sin^2 \frac{x}{2})}\]
welshfella
  • welshfella
sec (pi/6 - x) = 1 / (sqrt3/2 cos x + sin pi/6 sin x) = 2 / (sqrt 3 cos + sin x)
welshfella
  • welshfella
* I missed out the second step which is = 1 / (cos pi/6 cos x + sin pi/6 sin x)
anonymous
  • anonymous
\[\frac{ 2 }{\sqrt{3} \cos(x)+\sin(x)}=\sec(\pi/6-x)\] \[\frac{ 2 }{\sqrt{3} \cos(x)+\sin(x)}=\frac{ 1 }{ \cos{(\pi/6-x)} }\] using:cos(u-v)=cos u cos v+sin u sin v \[\frac{ 2 }{\sqrt{3} \cos(x)+\sin(x)}=\frac{ 1 }{ \cos{(\pi/6) \times \cos(x)}+\sin{(\pi/6) \times \sin(x)} }\]
anonymous
  • anonymous
i got that far ASAAD123 put i didn't know what to do after that
welshfella
  • welshfella
the denominator = cos (pi/6 - x)
welshfella
  • welshfella
and 1 / cos = sec
anonymous
  • anonymous
wouldn't you also change sin(pi/6) to 1/2
welshfella
  • welshfella
no its the sin of a compound angle (pi/6 - x) not sin pi/6
anonymous
  • anonymous
ok
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
so how do you get the sides to equal
welshfella
  • welshfella
i had to use sin pi/6 = 1/2 because i started with RHS and converted to LHS whereas ASAAD did the reverse
welshfella
  • welshfella
with these identities you choose one side and try to convert it to the other. This proves the identity.
anonymous
  • anonymous
ok
anonymous
  • anonymous
i think i understand it now
welshfella
  • welshfella
so you can use asaad's or mine. Either would do.
welshfella
  • welshfella
There is only more more step to prove IrishBoy's solution
anonymous
  • anonymous
i already got irish boy's solution

Looking for something else?

Not the answer you are looking for? Search for more explanations.