cosA+cos2A+cos6A+cos7A= 4cosA/2 cos5A/2 cos4A
(sinA-sinB)(sinA+sinB)= sin(A-B) sin(A+B)
(sin2A+sin5A-sin3A)/(cosA+1-2sin^2 2A) = 2sinA
I have no idea how to prove these identities. Could anyone help me saying any idea how to prove? Thanks
cosA+cos2A+cos6A+cos7A= 4cosA/2 cos5A/2 cos4A
(sinA-sinB)(sinA+sinB)= sin(A-B) sin(A+B)
(sin2A+sin5A-sin3A)/(cosA+1-2sin^2 2A) = 2sinA
I have no idea how to prove these identities. Could anyone help me saying any idea how to prove? Thanks
Hello rolisp
Welcome to Math Help Forum!In each of these questions you need to use the Sum-to-Product formulas (and in #2 and #3, the double-angle formulae) together with basic factorising and simplifying techniques.
I'll start you off in each case, and leave you to try to complete it.
1)Now factorise by taking out a common factor . Then use the same sum-to-product formula again, and you're there.
2)Now re-arrange, and use the identity (twice), to write the expression in terms of and .
3)
Factorise; cancel and you're there.
Let us know how you get on.
Grandad