Question

How do you evaluate `cos (arcsin (3/5) - arccos (1/2))`?

Answer 1

`cos [arcsin (3/5) - arccos (1/2)] =`

Explanation:

`sin x = 3/5 = 0.6`--> x = 36.87 and x = 180 - 36.87 = 143.13 deg

`cos y = 1/2` --> y = +- 60 deg

a. cos (x - y) = cos (143.13 - 60) = cos 83,13 `= 0.12`

b. cos (36.87 - 60) = cos (-23.23) = 0.92

Answer 2

An algebraic expression for this can be calculated using the properties of the right angled triangles with sides {3,4,5} and {1,sqrt(3),2} using the formula for `cos(alpha+beta)`.

Explanation:

`cos(alpha+beta)=cos(alpha)cos(beta)-sin(alpha)sin(beta)`

So

`cos(alpha-beta)=cos(alpha)cos(-beta)-sin(alpha)sin(-beta)`

`=cos(alpha)cos(beta)+sin(alpha)sin(beta)`

If `sin(alpha) = 3/5` and `0<=alpha<=pi/2` then

`cos(alpha) = 4/5`

If `cos(beta)=1/2` and `0<=beta<=pi/2` then

`sin(beta)=sqrt(3)/2`

Then

`cos(alpha-beta)`

`= cos(alpha)cos(beta)+sin(alpha)sin(beta)`

`= 4/5*1/2+3/5*sqrt(3)/2`

`= 4/10+(3sqrt(3))/10`

`= (4+3sqrt(3))/10`

`~= 0.9196`