Question

How do you find `cos67.5` using the half-angle identity?

Answer 1

`cos(67.5^@)=+(sqrt(2-sqrt2))/2`

Explanation:

The Half-Angle Identity is `1+costheta=2cos^2(theta/2)`

Taking, `theta=2xx67.5^@=135^@, 1+cos(135^@)=2cos^2(67.5^@)`

`:. 2cos^2(67.5^@)=1+cos(180^@-45^@)=1-cos45^@`

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

`:.cos^2(67.5^@)=(2-sqrt2)/4`

`:. cos(67.5^@)=+(sqrt(2-sqrt2))/2`, `+ve` sign, as, 67.5^@ is in the first Quadrant#

Answer 2

I got `sqrt(2 - sqrt2)/2` as well.

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Since `67.5^@` is half of `135^@`, let's work off of that. I remember it by starting from the `cos^2(x)` identity.

`cos^2(x) = (1 + cos(2x))/2`

`cos^2(x/2) = (1 + cosx)/2`

`:. cos(x/2) = pmsqrt((1 + cosx)/2)`

Since `67.5^@` is in quadrant `"I"`, where `0 < x < 90^@`, and `cos(x)` for `0 < x < 90^@` is positive, `cos(67.5^@) > 0`. Therefore:

`cos(135^@/2) = +sqrt((1 + cos(135^@))/2)`

`= sqrt((1 + cos((3pi)/4))/2)`

`= sqrt((1 - sqrt2/2)/2)`

`= sqrt((2/2 - sqrt2/2)/2) = sqrt(((2 - sqrt2)/2)/2)`

`= sqrt((2 - sqrt2)/4)`

`= color(blue)(sqrt(2 - sqrt2)/2 ~~ 0.3827)`