- #1

wei1006

- 6

- 0

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...