Question

If `cosA=24/25` and `cosB=3/5` and `A` lies in `Q1`, find `cos(A+B)` and `cos(A-B)`?

Answer 1

`cos(A+B)=44/125` and `cos(A-B)=4/5`

Explanation:

As angles A and B are acute angles all trigonometric ratios are positive, hence

as `cosA=24/25`,

`sinA=sqrt(1-(24/25)^2)=sqrt(1-576/625)=sqrt(49/625)=7/25`

and as `cosB=3/5`,

`sinB=sqrt(1-(3/5)^2)=sqrt(1-9/25)=sqrt(16/25)=4/5`

Hence `cos(A+B)=cosAcosB-sinAsinB`

= `24/25xx3/5-7/25xx4/5=(72-28)/125=44/125`

and `cos(A-B)=cosAcosB+sinAsinB`

= `24/25xx3/5+7/25xx4/5=(72+28)/125=100/125=4/5`

Answer 2

`cos(A+B)=44/125,cos(A-B)=4/5`

Explanation:

`"since A and B are both acute then we are in the "`
`"first quadrant"`

`"using the "color(blue)"trigonometric identities"`

`•color(white)(x)cos(A+-B)=cosAcosB∓sinAsinB`

`cosA=24/25`

`•color(white)(x)sinA=sqrt(1-cos^2A)`

`color(white)(xxxxxx)=sqrt(1-(24/25)^2)`

`color(white)(xxxxxx)=sqrt(1-576/625)`

`color(white)(xxxxxx)=sqrt(49/625)=7/25`

`cosB=3/5`

`rArrsinB=sqrt(1-9/25)=sqrt(16/25)=4/5`

`rArrcos(A+B)=(24/25xx3/5)-(7/25xx4/5)`

`color(white)(xxxxxxxxxx)=72/125-28/125=44/125`

`rArrcos(A-B)=72/125+28/125=100/125=4/5`