# How does $\sin \theta \times \sin (\theta - \phi)$ become $\frac{\cos \phi - \cos (2\theta-\phi)}{2}$

How does $\sin \theta \times \sin (\theta - \phi)$ become $\frac{\cos \phi - \cos (2\theta-\phi)}{2}$

• I recommend using something other than $\varnothing$ for an angle, or number. It by convention (that I have never seen broken) is the empty set. – Paul Plummer Mar 24 '15 at 7:28
• @PaulPlummer - the OP probably wanted to write $\phi$ here. – 299792458 Mar 24 '15 at 7:29
• @TheDarkSide I could tell (and was about to submit an edit before someone got to it before I did), just letting the author know for any future posts. – Paul Plummer Mar 24 '15 at 7:31
• @PaulPlummer Yeah.will do.Thanks – techno Mar 24 '15 at 9:19
• – lab bhattacharjee Mar 24 '15 at 15:58

Directly expanding the LHS would be a difficult way of going about it, since there would have to be so many simplifications. Instead, proceed as follows:

Use the identities

$$\cos (A \pm B) = \cos A \cos B \mp \sin A \sin B$$

to arrive at

$$\sin A \sin B = \left(\cos (A-B) - \cos (A+B) \right)/2$$

Then, your answer follows in one step. Note that cosine function is even under parity, so that $\cos(-\theta) = \cos(\theta)$. SUBSTITUTE FORMULE FOR SIN(A-B) AND (SIN^2θ) AND LASTLY GET THE EQUATION IN FORM OF COS(A-B) :)

• Don't use all caps. Also, it is frowned upon to post a photo in most instances (and I think this is one of those). You may want to learn some MathJax so that you can provide answer that are easier to read and better formated. A tutorial can be found here – Paul Plummer Mar 24 '15 at 7:45
• Thx!!a lot!! the link was very helpful... @PaulPlummer – Suhaas Mar 24 '15 at 9:08
• Thanks for writing this,i have upvoted it.But i will select the first answer as its simple. – techno Mar 24 '15 at 15:41