Question

If `sin(theta) = 3/5` and `theta` is in Q1, then what is `sin(90^@+theta)` ?

Answer 1

`4/5`

Explanation:

Note that:

`cos^2 theta + sin^2 theta = 1`

So:

`cos theta = +-sqrt(1-sin^2 theta)`

`color(white)(cos theta) = +-sqrt(1-(3/5)^2)`

`color(white)(cos theta) = +-sqrt((25-9)/25)`

`color(white)(cos theta) = +-sqrt(16/25)`

`color(white)(cos theta) = +-4/5`

We can identify the correct sign as `+` since `theta` is in Q1.

So:

`cos theta = 4/5`

The sum of angles formula for `sin` tells us:

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

Putting `alpha = 90^@` and `beta = theta` we find:

`sin(90^@+theta) = sin(90^@)cos(theta)+sin(theta)cos(90^@)`

`sin(90^@+theta) = 1*cos(theta)+sin(theta)*0`

`sin(90^@+theta) = cos(theta)`

`sin(90^@+theta) = 4/5`

Note that in passing we have shown:

`sin(90^@+theta) = cos(theta)`

for any `theta`.

That is, `cos(theta)` is the same as `sin(theta)` shifted by `90^@`.