Question

How do you use the half-angle identity to find the exact value of cos(17π/12)?

original question: "Use a half-angle formula to find the exact value of cos 17π/12"

Answer 1

As demonstrated below

Explanation:

I derive formulae as I typically do and I hope you find it helpful in spite of being a little long.

First, we find the half-angle formula for the cosine. We know that
`cos(2x) = cos^2(x) - sin^2(x) = cos^2(x) - (1 - cos^2(x)) = 2cos^2(x) - 1` so solving for `cos(x)` we get
`cos(x) = +- sqrt((cos(2x) + 1)/2)`,
where we have to determine the sign later. Replacing `x` with `x/2` we get a half-angle formula
`cos(x/2) = +- sqrt((cos(x) + 1)/2)`.

If we let `x = (17 pi)/6`, then we get that
`cos((17 pi) /12) = +- sqrt((cos((17 pi)/6) + 1)/2)`.

Let us compute `cos((17 pi)/6)` by observing that `(17 pi)/6 = (18 pi)/6 - pi/6 = 3pi - pi/6`. Rewriting fractions of `pi` to a sum of an integer multiple of `pi` and a smaller fraction of `pi` is often useful for finding exact values in trigonometric tables.

Using that
`cos(2pi + y) = cos(y)`,
`cos(pi + y) = -cos(y)`,
and
`cos(-y) = cos(y)`
we find that
`cos((17pi)/6) = cos((18pi)/6 - pi/6) = cos(2pi + pi - pi/6) = -cos(-pi/6) = -cos(pi/6),`
which we can find in standard tables (or the unit circle) to be equal to `-sqrt(3)/2`. Plugging into our half-angle formula we get that
`cos((17 pi) /12) = +- sqrt(-sqrt(3)/2 + 1)/2`.

What remains now is to figure out the sign of `cos((17 pi)/12)`. We find that the angle is in third quadrant by observing that
`pi<(17 pi)/12<(18 pi)/12=(3 pi)/2`.
There, the cosine is negative (look up in a table or book, or look at the unit circle).

Therefore, we must choose the negative sign, and conclude that
`cos((17 pi) /12) = - 1/2sqrt(-sqrt(3)/2 + 1)`.

Answer 2

`cos ((17pi)/12) = - sqrt(2 - sqrt3)/2`

Explanation:

Use trig table and unit circle:
`cos ((17pi)/12) = cos ((-7pi)/12 + 2pi) = cos ((-7pi)/12) = cos ((7pi)/12)`
`= cos (pi/12 + pi/2) = - sin (pi/2)`
Find sin (pi/12) by applying trig identity -->
`2sin^2 a = 1 - cos 2a`
In this case:
`2sin^2 (pi/12) = 1 - cos (pi/6) = 1 - sqrt3/2 = (2 - sqrt3)/2`
`sin^2 (pi/12) = (2 - sqrt3)/4`
`sin (pi/12) = +- sqrt(2 - sqrt3)/2`
Since `sin (pi/12)` is positive, take the positive value.
Finally,
`cos ((17pi)/12) = - sin (pi/12) = - sqrt(2 - sqrt3)/2`