Question

How do you find the exact functional value sin(arcsin3/5 + arccos3/5) using the cosine sum or difference identity?

Answer 1

The value is `1`

Explanation:

We have that

`sin(arccos(x)) = cos(arcsin(x)) = sqrt(1 - x^2)`

Set
`arcsin(3/5) = a , arccos(3/5) = b`
Hence we have that

#sin(a + b) = sin(a)cos(b) + sin(b)cos(a) = sin(arcsin(3/5)) * cos(arccos(3/5)) + sin(arccos(3/5)) * cos(arcsin(3/5)) = (3/5) * (3/5) + sqrt(1 - (9/25)) * sqrt(1 - (9/25)) = (9/25) + 1 - 9/25 = 1 #

Answer 2

Find the value of `sin(arcsin (3/5) + arccos (3/5))`

Ans: 1

Explanation:

Call `arcsin (3/5) = x` and `arccos (3/5) = y.`
Use the trig identity: `sin x = cos(pi/2 - x)`
We have `3/5 = sin x = cos y = cos (pi/2 - x)`.
This means -->`(x + y) = pi/2` (complement arcs)
Therefor, `sin (x + y) = sin (pi/2) = 1`