How is sinx/(2sin(x/2)) = cos(x/2)?

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SUMMARY

The discussion focuses on proving the identity sin(x)/(2sin(x/2)) = cos(x/2) using trigonometric identities. Participants emphasize the importance of applying the double angle identity for sine, specifically rewriting sin(x) as sin(2(x/2)). Additionally, a second identity involving sin(x) and cos(x/(2^(n+1))) is presented, requiring similar manipulation of sine functions. The conversation highlights the necessity of demonstrating work in homework help forums to receive assistance.

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  • Understanding of trigonometric identities, particularly the double angle identity.
  • Familiarity with the sine and cosine functions.
  • Basic algebraic manipulation skills.
  • Knowledge of how to apply trigonometric identities in proofs.
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  • Study the double angle identity for sine and its applications.
  • Learn how to manipulate trigonometric functions for proofs.
  • Explore advanced trigonometric identities and their derivations.
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Homework Statement


Prove sinx/(2sin(x/2)) = cos(x/2)
and
(sin(x) cos(x/2^(n+1)))/(2^n(sinx/2^n)) = sinx/(2^(n+1)sin(x/(2^n+1)))


Homework Equations





The Attempt at a Solution

 
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mr0no said:

Homework Statement


Prove sinx/(2sin(x/2)) = cos(x/2)
and
(sin(x) cos(x/2^(n+1)))/(2^n(sinx/2^n)) = sinx/(2^(n+1)sin(x/(2^n+1)))


Homework Equations





The Attempt at a Solution


For the first, how do you expand \sin(2x) into terms of sin(x) and cos(x)? Now apply that to the numerator sin(x).

For the second, again apply the same rule.
 
mr0no said:

Homework Statement


Prove sinx/(2sin(x/2)) = cos(x/2)
and
(sin(x) cos(x/2^(n+1)))/(2^n(sinx/2^n)) = sinx/(2^(n+1)sin(x/(2^n+1)))

Homework Equations



The Attempt at a Solution

Hello mr0no. Welcome to PF !

According to the rules for homework help in this Forum, you need to show your work before we can help.


Since you're new here, I'll give you a hint.

For the first one use the double angle identity to write sin(x) in a different manner, by looking at sin(x) as sin(2(x/2)) .
 
Thanks for giving me a tip instead of solving the whole thing for me. That's just what I need. I guess I will be revising trig identities tomorrow :)
 

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