Asked by Karla
                Can you please help me verify these?
>sin7x-sin5x= tanx(cos7x-cos5x)
>(1-cosx)cot(1/2x) = sinx
>1+tanx tan(x/2) = secx
Its very confusing because of all the double-half angle formulas >_<
            
        >sin7x-sin5x= tanx(cos7x-cos5x)
>(1-cosx)cot(1/2x) = sinx
>1+tanx tan(x/2) = secx
Its very confusing because of all the double-half angle formulas >_<
Answers
                    Answered by
            Reiny
            
    I tried your #1 with x=10° and it did not work
so it is not even an identity
#2 and #3.
you will have to remember that
cos 2A = 1 - 2sin^2 A , and sin 2A = 2sinAcosA
or
cosx = 1 - 2sin^2 (x/2) , and sinx = 2sin(x/2)cos(x/2)
    
so it is not even an identity
#2 and #3.
you will have to remember that
cos 2A = 1 - 2sin^2 A , and sin 2A = 2sinAcosA
or
cosx = 1 - 2sin^2 (x/2) , and sinx = 2sin(x/2)cos(x/2)
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