Question

How do you evaluate `sin (arccos.5 + arcsin .6)`?

Answer 1

Evaluate sin [arccos 0.5 + arcsin 0.6]

Ans: 0,99

Explanation:

Using calculator:
cos x = 0.5 --> arc x = 60 deg
sin x = 0.6 --> arc x = 36.87 deg
sin (60 + 36.87) = sin 96.87 = 0.99

Answer 2

`sin(arccos 0.5 + arcsin 0.6) = 0.3 + 0.8sqrt0.75`

Explanation:

If we want to do this without a calculator or trig tables, use the following:

`arccos 0.5` is some `alpha` in `[0, pi]` with `cos alpha = 0.5`.
We will need `sin alpha` so we note that with `alpha` in `[0, pi]`, we have `sin alpha` is positive.

Therefore `sin alpha = sqrt(1-cos^2 alpha) = sqrt 0.75`

By similar reasoning, `arcsin 0.6` is some `beta in [-pi/2, pi/2]` with `sin beta = 0.6` and `cos beta = sqrt(1-0.36) = sqrt (0.64) = 0.8`

We have been asked to find `sin(alpha + beta)`.

Use

`sin(alpha + beta) = sin alpha cos beta + cos alpha sin beta` and the values above to get:

`sin(alpha + beta) = (sqrt0.75)(0.8)+(0.5)(0.6)`

`= 0.3 + 0.8sqrt0.75`