Question

How do you find the exact values of sin 15 degrees using the half angle formula?

Answer 1

I found: `sin(15°)=0.258`

Explanation:

Using the Half Angle Formula:
`color(red)(sin^2(x)=1/2[1-cos(2x)])`

with `x=15°` and `2x=30°`

you get:

`sin^2(15°)=1/2[1-cos(30°)]`

knowing that: `cos(30°)=sqrt(3)/2`:

`sin^2(15°)=1/2[1-sqrt(3)/2]`
`sin^2(15°)=(2-sqrt(3))/4=0.067`

So:
`sin(15°)=+-sqrt(0.067)=+-0.258`
We choose the positive one.

Answer 2

`sin(15^@)=sqrt(2-sqrt(3))/2=(sqrt(6)-sqrt(2))/4`
See below.

Explanation:

Find exact value of `sin(15^@)` with half-angle formula.

Consider the half-angle formula for sine: `sin(theta/2)=sqrt((1-cosx)/2)`

Since we know that 15 is half of 30, we can plug `30^@` in as `theta` and simplify:
`sin(15^@)=sin(30^@/2)=sqrt((1-cos(30^@))/2)`
`=sqrt((1-sqrt(3)/2)/2)`
`=sqrt(((2-sqrt(3))/2)/2)`
`=sqrt((2-sqrt(3))/4)`
or see slightly more advanced method to remove nested root (at the bottom)
`=sqrt(2-sqrt(3))/2`
which is our answer, but has a nested root
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Removing nested root from answer:
Let us multiply both the numerator and denominator inside the square root by `2`:
`=sqrt((2-sqrt(3))/4*2/2)`
`=sqrt((4-2sqrt(3))/8)`

Now we can write `4-2sqrt(3)` as a square in the numerator:
`=sqrt((sqrt(3)-1)^2/8)`

We can take out a `sqrt(3)-1` from the numerator and a `2` from the denominator:
`=((sqrt(3)-1)/2)*1/sqrt(2)`
`=(sqrt(3)-1)/(2sqrt(2))`

We can rationalize the denominator:
`=(sqrt(2)(sqrt(3)-1))/4`
`=(sqrt(6)-sqrt(2))/4`

which is our answer without a nested root.

Answer 3

`(sqrt6-sqrt2)/4`

Explanation:

Use

`sin(A-B) = sinA*cosB - SinB*cosA`

So

`sin 15 = sin(45-30) = sin45*cos30 - sin30*cos45`

We know

`sin45 = cos45 = (sqrt2)/2`

We also know

`cos30 = (sqrt3)/2`

`sin30 = 1/2`

Plugging in the values, we get

`sin15 = (sqrt2)/2*((sqrt3)/2-1/2)`

Simplifying, we get

`(sqrt6-sqrt2)/4`