Question

Question #f1b58

Answer 1

see explanation.

Explanation:

Using the `color(blue)"product to sum identity"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(cosx cosy=1/2[cos(x+y)+cos(x-y)])color(white)(2/2)|)))`

`"here "x=(alpha+beta),y=(alpha-beta)`

`rArrx+y=alpha+beta+alpha-beta=2alpha`

`rArrx-y=alpha+beta-alpha+beta=2beta`

`rArrcos(alpha+beta)cos(alpha-beta)=1/2[cos2alpha+cos2beta]`

Using the `color(blue)"Double angle identities"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(cos2x=1-2sin^2x=2cos^2x-1)color(white)(2/2)|)))`

`rArr1/2[cos2alpha+2cosbeta]`

`=1/2[1-2sin^2alpha+2cos^2beta-1]`

`=cos^2beta-sin^2alpha`

`"Since left side "=" right side"rArr"verified"`

Answer 2

see below

Explanation:

One way is as follows: start with the compound angle expansions

`cos(alpha+beta)cos(alpha-beta)`

`=(cosalphacosbeta-sinalphasinbeta)(cosalphacosbeta+sinalphasinbeta)`

this is the difference of squares , hence:

`=cos^2alphacos^2beta-sin^2alphasin^2beta`

now `sin^2x+cos^2x=1`

so `cos^2alpha=1-sin^2alpha`

& `sin^2beta=1-cos^2beta`

we have

`=(1-sin^2alpha)cos^2beta-sin^2alpha(1-cos^2beta)`

multiply out and simplify

`=cos^2beta-sin^2alphacos^2beta-sin^2alpha+sin^2alphacos^2beta`

`=cos^2betacancel(-sin^2alphacos^2beta)-sin^2alphacancel(+sin^2alphacos^2beta)`

`=cos^2beta-sin^2alpha" as required"`