Question

If `sintheta_1+sintheta_2+sintheta_3=3`, what is the value of `costheta_1+costheta_2+costheta_3`?

Answer 1

`costheta_1+costheta_2+costheta_3=0`

Explanation:

Value of sine ratio ranges from `-1` to `1` i.e. `[-1,1]`

Now as `sintheta_1+sintheta_2+sintheta_3=3`

We must have each of them as `1`

and hence `theta_1=theta_2=theta_3=90^@`

and as `cos90^@=0`

`costheta_1+costheta_2+costheta_3=0`

Answer 2

Given

`sin theta_1 + sin theta_2 + sin theta_3=3`

`=>1-sin theta_1 +1- sin theta_2 +1- sin theta_3=0`

`=>1-2sin (theta_1/2)cos(theta_1/2)+1- 2sin (theta_2/2)cos(theta_2/2) +1- 2sin (theta_3/2)cos(theta_3/2)=0`

`=>sin^2 (theta_1/2)+cos^2(theta_1/2)-2sin (theta_1/2)cos(theta_1/2)+sin^2 (theta_2/2)+cos^2(theta_2/2)- 2sin (theta_2/2)cos(theta_2/2) +sin^2 (theta_3/2)+cos^2(theta_3/2)- 2sin (theta_3/2)cos(theta_3/2)=0`

`=>(sin (theta_1/2)-cos(theta_1/2))^2+(sin (theta_2/2)-cos(theta_2/2))^2 +(sin (theta_3/2)-cos(theta_3/2))^2=0`

Sum of three squared quantities being zero each quantity should be zero

So

`sin (theta_1/2)-cos(theta_1/2)=0`

`=>tan (theta_1/2)=1=tan(pi/4)`

Hence `theta_1=pi/2`
Similarly

`theta_2=pi/2`

and

`theta_3=pi/2`

then the value of `cos theta_1+ cos theta_2 + cos theta _3`

`=cos (pi/2)+ cos (pi/2) + cos(pi/2)=0+0+0=0`