Establish each identity.
(1) [sin(4x) - sin(8x)]/[cos(4x) - cos(8x)] = -cot(6x)
(2) (cos A + cos B)/(cos A - cos B) =
[-cot(A + B)/2]*[cot(A - B)/2]
Hi,
$\displaystyle \sin(a)-\sin(b)=2 \cos \left(\frac{a+b}{2}\right) \sin \left(\frac{a-b}{2}\right)$ (1)
$\displaystyle \cos(a)-\cos(b)=-2 \sin \left(\frac{a+b}{2}\right) \sin \left(\frac{a-b}{2}\right)$ (2)
$\displaystyle \cos(a)+\cos(b)=2 \cos \left(\frac{a+b}{2}\right) \cos \left(\frac{a-b}{2}\right)$ (3)
(1) and (2)
(2) and (3)(2) (cos A + cos B)/(cos A - cos B) =
[-cot(A + B)/2]*[cot(A - B)/2]
Please learn these formulae ! Try to search before asking, otherwise it kills all the working of thinking !
magentarita, my question to you is simplify "Why can you do the most basic mathematical operation? I think that you are simply taking advantage of this site. Are you simply getting all of your problems done you? IS THAT CHEATING?
If it is no cheating, please explain why it is not cheating
Come on magentarita!this is a very very basic question.may be you don't wanna learn maths,you just wanna copy for your homework(your question indicates that).if presently you are not having any book of trigonometry than this question is fine but if you are having some book then you are not at all opening it(your question shows).
OK i will give you the complete solution but tell what problem you faced in solving.
But hey you may ask any question,feel free for that.
But what problem one can face in such an easy problem,i would like to know it for sure
For what it's worth, I think this particular OP is lacking confidence rather than ethics. Attempting to overcome maths phobia (and/or dislike) may not be too strong a term, perhaps ......
(*Shrug* But Barnum always had a big smile on his face when he saw me coming along. "This way to the egress, good sir" ......)
Anyhow, I think the moral of the story is that all posters should try and show some effort/attempt at solving the questions they post.