Question

How do you find the exact value of sin67.5 degrees?

Answer 1

We could use the half-angle identity:
`sin(1/2x)= +- sqrt((1-cosx)/2)`

Explanation:

`67.5^2 = 1/2(135^@)` (Multiply `67.5 xx 2`.)

The formula doesn't tell us whether `sin 67.5^@` is positive or negative, but, since it is an acute angle we know that the sine is positive. (Be careful of the difference between "sign" and "sine").

We also need `cos135^@`. (That is the special angle that is `45^@` in Quadrant II.)

`cos135^@ = -sqrt2/2`.

Now use the formula ans simplify:

`sin 67.5^@ = sin(1/2(135^@)) = sqrt((1-cos135^@)/2)`

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

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

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

Answer 2

Find sin 67.5

Explanation:

Call sin 67.5 = sin x

`cos 2x = cos 135 = - (sqrt2)/2 = 1 - 2sin^2 x`

`2sin^2 x = 1 + (sqrt2)/2 = (2 + sqrt2)/2`

`sin^2 x = (2 + sqrt2)/4`

`sin x = sin 67.5 = (sqrt(2 + sqrt2))/2`