Question

If `sinA+sinB=x` and `cosA+cosB=y` then show that `tan((A-B)/2)=+-sqrt((4-x^2-y^2)/(x^2+y^2))`?

Answer 1

See the answer below...

Explanation:

• `sinA+sinB=x`
  `=>2 cdot sin((A+B)/2)cdot cos((A-B)/2)=x" "...(1)`

• `cosA+cosB=y`
  `=>2cdot cos((A+B)/2) cdot cos((A-B)/2)=y" "...(2)`

  Now, we have to remove the term of sine and cosine value of #" "((A+B)/2)" "#to get the value of `tan((A-B)/2)`.

From 1st equation,

`=>4 cdot sin^2((A+B)/2)cdot cos^2((A-B)/2)=x^2`
`=>4 cdot cos^2((A-B)/2)cdot {1-cos^2((A+B)/2)}=x^2" "...(3)`

Similarly, from 2nd equation,

`=>4 cdot cos^2((A+B)/2) cdot cos^2((A-B)/2)=y^2`
`=>4 cdot cos^2((A+B)/2) cdot cos^2((A-B)/2)=y^2" "...(4)`

From 3rd and 4th equation, we get

`=>4 cdot cos^2((A-B)/2)cdot {1-y^2/(4 cdot cos^2((A-B)/2)}}=x^2`

`=>4 cdot cos^2((A-B)/2)-y^2 =x^2`

`=>4 cdot cos^2((A-B)/2) =x^2+y^2`

`=>cos^2((A-B)/2)=(x^2+y^2)/4`

`=>sec^2((A-B)/2)=4/(x^2+y^2)`

`=>1+tan^2((A-B)/2)=4/(x^2+y^2)`

`=>tan^2((A-B)/2)=4/(x^2+y^2)-1`

`=>tan^2((A-B)/2)=(4-x^2-y^2)/(x^2+y^2)`

`=>tan((A-B)/2)=+-sqrt((4-x^2-y^2)/(x^2+y^2)`

Hope it helps..
Thank you...

Answer 2

Proof is done considering squaring both the sides, and verified

Explanation:

Given:
`sinA+sinB=x`
`cosA+cosB=y`

To show that:

`tan((A-B)/2)=+-sqrt((4-x^2-y^2)/(x^2+y^2))`

Squaring both sides,

`tan^2((A-B)/2)=(4-x^2-y^2)/(x^2+y^2)`

RHS

`=(4-(sinA+sinB)^2-(cosA+cosB)^2)/((sinA+sinB)^2+(cosA+cosB)^2)`

`=(4-(sin^2A+2sinAsinB+sin^2B)-(cos^2A+2cosAcosB+cos^2B))/(sin^2A+2sinAsinB+sin^2B+cos^2A+2cosAcosB+cos^2B`

`=(4-(sin^2A+cos^2A+sin^2B+cos^2B+2(cosAcosB+sinAsinB)))/(sin^2A+cos^2A+sin^2B+cos^2B+2(cosAcosB+sinAsinB))`

`sin^2A+cos^2A=1`
`sin^2B+cos^2B=1`
`cosAcosB+sinAsinB=cos(A-B)`

`=(4-(1+1+2cos(A-B)))/(1+1+2cos(A-B))`

`=(4-1-1-2cos(A-B))/(1+1+2cos(A-B)`

`(2-2cos(A-B))/(2+2cos(A-B))`

`(2(1-cos(A-B)))/(2(1+cos(A-B)))`

`=(1-cos(A-B))/(1+cos(A-B))`

`=(2sin^2((A-B)/2))/(2cos^2((A-B)/2))`

`=(sin^2((A-B)/2))/(cos^2((A-B)/2))`

`=tan^2((A-B)/2)`

=LHS

Hence, proved