1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Trig half-angle identities?

  1. Oct 6, 2012 #1
    Apparently our professor expects us to know these half-angle identities

    idents03.gif
    (http://www.purplemath.com/modules/idents.htm)

    Without going through them in class or us learning them in high school..

    Can somebody explain how these were derived? Does the derivation come from the angle-sum and difference identities?
     
    Last edited: Oct 6, 2012
  2. jcsd
  3. Oct 6, 2012 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi Nikitin! :smile:

    Your professor is right, you do need to learn these.
    Yes, have a go! :smile:
     
  4. Oct 6, 2012 #3
    thx. With a bit of effort it wasn't that hard :)
     
  5. Oct 6, 2012 #4

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    These are usually stated a bit differently:

    For example, for sin2(x) = ½[1 – cos(2x)],

    Taking the square root of both sides of the above identity, and letting θ = 2x, we have,

    [itex]\displaystyle \sin(\theta/2)=\pm\sqrt{\frac{1-\cos(\theta)}{2}}\,,[/itex] where the ± sign indicates that you choose the correct sign depending upon the quadrant in which θ/2 lies.

    Similar results hold for the other half-angle identities.
     
  6. Oct 6, 2012 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The fundamental identities are sin^2(a)+ cos^2(a)= 1, sin(a+ b)= sin(a)cos(b)+ cos(a)sin(b) and cos(a+ b)= cos(a)cos(b)- sin(a)sin(b). How you show those depends on exactly how you have defined "sine" and "cosine".

    But from those cos(2a)= cos(a+ a)= cos^2(a)- sin^2(a). Of course, then, cos(2a)= cos^2(a)- (1- cos^2(a)= 2cos^2(a)- 1 and cos(2a)= (1- sin^2(a))- sin^2(a)= 1- 2sin^2(a).

    From cos(2a)= 2cos^2(a)- 1, we have cos^2(a)= (1/2)(cos(2a)+ 1) so that [itex]cos(a)= \pm\sqrt{(1/2)(cos(2a)+ 1)}[/itex] and, setting b= 2a, [itex]cos(b/2)= \pm\sqrt{(1/2)(cos(b)+ 1)}[/itex].

    From cos(2a)= 1- 2sin^2(a), we have sin^2(a)= (1/2)(1- cos(2a) so that [itex]sin(a)= \pm\sqrt{(1/2)(1- sin(2a)}[itex] and, setting b= 2a, [itex]sin(b/2)= \pm\sqrt{(1/2)(cos(b)+ 1)}[/itex].

    Whether we use + or - for a given b depends on the quadrant b/2 is in which is NOT given by cos(b).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Trig half-angle identities?
  1. Trig identity (Replies: 2)

  2. Trig identity (Replies: 19)

  3. Trig identity (Replies: 9)

  4. Trig Identities (Replies: 8)

  5. Trig identity (Replies: 2)

Loading...