Question

How do you verify the identity `(cos3beta)/cosbeta=1-4sin^2beta`?

Answer 1

`cos(3beta) = cos(2beta + beta)`.

Use the sum formula `cos(A + B) = cosAcosB - sinAsinB` to expand:

LHS:

`(cos2betacosbeta - sin2betasinbeta)/cosbeta`

Expand using the identities `cos2x = 2cos^2x - 1` and `sin2x = 2sinxcosx`:

`((2cos^2beta- 1)cosbeta - (2sinbetacosbeta)sin beta)/cosbeta`

`(2cos^3beta - cosbeta - 2sin^2betacosbeta)/cosbeta`

Use `sin^2x + cos^2x = 1`:

`(2cos^3beta - cosbeta - 2(1 - cos^2beta)cosbeta)/cosbeta`

`(2cos^3beta - cosbeta - 2(cosbeta - cos^3beta))/cosbeta`

`(2cos^3beta - cosbeta - 2cosbeta + 2cos^3beta)/cosbeta`

`(4cos^3beta - 3cosbeta)/cosbeta`

`(cosbeta(4cos^2beta - 3))/cosbeta`

`4cos^2beta - 3`

Switch into sine now using `cos^2x = 1 - sin^2x`:

`4(1 - sin^2beta) - 3`

`4 - 4sin^2beta - 3`

`1 - 4sin^2beta`

Since the `LHS` equals the `RHS`, we are done here.

Hopefully this helps!

Answer 2

See the Proof in the Explanation Section.

Explanation:

We will need `cos2beta=1-2sin^2beta`

`cos3beta=ul(cos3beta+cosbeta)-cosbeta.................(1)`

Since, `cosC+cosD=2cos((C+D)/2)cos((C-D)/2)`

`:.," from "(1), cos3beta=2cos2betacosbeta-cosbeta,`

`=(cosbeta)(2cos2beta-1),`

`=(cosbeta){2(1-2sin^2beta)-1}, i.e.,`

`cos3beta=(cosbeta)(1-4sin^2beta),`

`rArr (cos3beta)/cosbeta=1-4sin^2beta`

Enjoy Maths.!