Question

Prove that cos 3A = cos (2A+A)

cos 3A = 4 cos³A – 3cosA..... Where from the cos A ....does sin A * cos A give cos A??


Cos 3A = cos (2A +A)
Cos 2A cos A - Sin2A sinA
(2cos²A-1)cosA-(2sinA cos A) sinA
2cos³A - cos A - 2 sin²A cos A
2 cos³A-cos A - 2(1-cos²A)cos A
2 cos³A - cos A - 2cos A + 2 cos³A

Cos3A = 4cos³A-3cos A​

Answer 1

Not sure what the question is, but I guess it's to prove that

`\cos3A=4\cos^3A-3\cos A`

Expand the left side as

`\cos3A=\cos(2A+A)=\cos2A\cos A-\sin2A\sin A`

and use the double angle identities to write

`\cos3A=(\cos^2A-\sin^2A)\cos A-2\sin^2A\cos A`

`\cos3A=\cos^3A-3\sin^2A\cos A`

Recall the Pythagorean identity:

`\cos3A=\cos^3A-3(1-\cos^2A)\cos A`

`\cos3A=\cos^3A-3\cos A+3\cos^3A`

`\implies\cos3A=4\cos^3A-3\cos A`

as required.