Further Trigonometry Identity (Proving question)

[wei1006]

1) Question:
Show that (sin3A-sinA)/(cosA+cos3A)=tanA

2) Relevant equations:
tan A=sinA/cos A
1+tan^2A=sec^A
cot A=1/tanA
cot A=cos A/sinA
sin^2A+cos^2A=1
secA=1/cos A
cosecA=1/sinA
1+cosec^2A= cot^2A
sin2A=2sinAcosA
cos2A=1-2sin^2A=cos^2A-sin^2A=2cos^A-1
tan2A=(2tanA)/1-tan^2A

3)Attempt:
(sin3A-sinA)/(cos A+cos3A)
=(2sin3/2Acos3/2A)-(sinAcosA)/(cos^2(1/2)A-sin^2(1/2)A)+(cos^2(3/2)A-sin^2(3/2)A)
I tried expanding but ended up confusing myself...

 

[Jimmy Johnson]

Try using the expressions :

sin(x) - sin(y) = 2cos((x+y)/2)sin((x-y/2)

cos(x) + cos(y) = 2cos((x+y)/2)cos((x-y)/2)

find identities here : http://www.purplemath.com/modules/idents.htm