Question

Sin(A+B)=?.

Answer 1

`a)` `63/65`, `b)` `33/65`, `c)` `63/16`, `d)` `33/56`, `e)` `1`, `f)` `1`

Explanation:

.

`sin^2A+cos^2A=1, :.sinA=sqrt(1-cos^2A)=sqrt(1-9/25)=+-4/5`

Since the problem states that `pi < A <(3pi)/2`, Angle `A` is in the third quadrant. In that quadrant, `sin` values are negative. Therefore, `sinA=-4/5`

Similarly, `sinB=sqrt(1-cos^2B)=sqrt(1-144/169)=+-5/13`

Since angle `B` is also in the third quadrant:

`sinB=-5/13`

`color(red)(a))` `sin(A+B)=sinAcosB+cosAsinB=(-4/5)(-12/13)+(-3/5)(-5/13)=48/65+15/65=63/65`

`color(red)(b))` `sin(A-B)=sinAcosB-cosAsinB=48/65-15/65=33/65`

`cos(A+B)=cosAcosB-sinAsinB=(-3/5)(-12/13)-(-4/5)(-5/13)=36/65-20/65=16/65`

`cos(A-B)=cosAcosB+sinAsinB=36/65+20/65=56/65`

`color(red)(c))` `tan(A+B)=sin(A+B)/cos(A+B)=(63/65)/(16/65)=63/16`

`color(red)(d))` `tan(A-B)=sin(A-B)/cos(A-B)=(33/65)/(56/65)=33/56`

`color(red)(e))` Since angle `(A+B)` has both `sin` and `cos` as positive values, it is in quadrant `1`.

`color(red)(f))` Since angle `(A-B)` has both `sin` and `cos` as positive values, it is in quadrant `1`.