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marketa · Oct 27, 2011, 07:24 AM

The task is to prove that they are equal ;)

ebaines · Oct 27, 2011, 07:44 AM

I think what you mean is this:

$ \sin(45+x) \sin (45-x) = \frac 1 2 - sin^2x $

This comes directly from substituting the identities:

$ \sin (a+b) = \sin (a) \cos (b) + \sin (b) \cos (a)\\ \sin (a-b) = \sin (a) \cos (b) - \sin (b) \cos (a)$

You will also use: $\cos ^2 x = 1 - \sin^2 x $

Post back to show us your progress, and if you're still stuck we can help you along.

marketa · Oct 27, 2011, 07:53 AM

h,I at first thank you very much for trying to help me ;) but the first part it should be 45 degrees and than the last member should be 1/2- sin2x (not squared) and I need to prove that they are equal so I need the whole modifying procedure