Question

Given that (sin α =24/25), with α in QI, and sin(α + β) = 12/13, with α + β in QII, find sin β and cos β. ?

I know that I need to use a Sum-to-product identity in order to solve this but I simply cannot figure out how. Thank you in advance for your help.

Answer 1

Given
`sinalpha=24/25andsin(alpha+beta)=12/13`

`alphainQ1->alpha " is acute"`

`cosalpha=sqrt(1-sin^2alpha)=sqrt(1-24^2/25^2)=7/25`

and

`alpha+betainQ2->(alpha+beta)" is obtuse"`

`cos(alpha+beta)=-sqrt(1-sin^2(alpha+beta))`
`=-sqrt(1-12^2/13^2)`

`=-5/13`

Now

`sinalphacosbeta+cosalpha sinbeta=sin(alpha+beta)=12/13`

`=>24/25cosbeta+7/25 sinbeta=12/13......[1]`

Again

`cosalphacosbeta-sinalphasinbeta=cos(alpha+beta)=-5/13`

`=>7/25cosbeta-24/25sinbeta=-5/13......[2]`

Multiplying [1]by 24 and [2] by 7 and then adding we get

`(24^2/25+7^2/25)cosbeta=(24*12-7*5)/13`
`=>25cosbeta=253/13`

`=>cosbeta=253/325`

Multiplying [1]by 7 and [2] by 24 and then subtracting we get

`(7^2/25+ 24^2/25)sinbeta=(12*7+24*5)/13`
`=>25sinbeta=204/13`

`=>sinbeta=204/325`

Answer 2

`sinbeta=204/325 and cosbeta=253/325`.

Explanation:

`sinalpha=24/25 :. cos^2alpha=1-sin^2alpha=1-(24/25)^2=(7/25)^2.`

`:. cosalpha=+-7/25," reading with, "alpha in QI`, we have,

`cosalpha=+7/25`,

Similarly, `sin(alpha+beta)=12/13, (alpha+beta) in QII, gives,`

`cos(alpha+beta)=-5/13`.

To sum up, we have,

`sin(alpha+beta)=12/13, cos(alpha+beta)=-5/13, cosalpha=7/25, and,`

`sinalpha=24/25`.

Therefore, `sinbeta=sin{(alpha+beta)-alpha}`,

`=sin(alpha+beta)cosalpha-cos(alpha+beta)sinalpha`,

`=12/12*7/25-(-5/13)24/25`.

`rArr sinbeta=204/325`.

On the similar lines,

`cosbeta=-5/13*7/25+12/13*24/25=253/325`