How do you use the half-angle identity to find the exact value of sin67.5°?

Question

How do you use the half-angle identity to find the exact value of sin67.5°?

1 Answer

Impact of this question

5725 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-17T22:51:49

Find ${sin 67}^{\circ} 5$ $sin 67^@5$

Ans: sqrt(2 + sqrt2)/2

Explanation:

Call sin (67.5) = sin t
cos 2t = cos 135 = -sqrt2/2 (Trig Table of Special Arcs).
Apply the trig identity: $cos 2 t = 1 - 2 {sin}^{2} t .$ $cos 2t = 1 - 2sin^2 t.$
$- \frac{\sqrt{2}}{2} = 1 - 2 {sin}^{2} t$ $-sqrt2/2 = 1 - 2sin^2 t$
$2 {sin}^{2} t = 1 + \sqrt{2} = \frac{2 + \sqrt{2}}{2}$ $2sin^2 t = 1 + sqrt2 = (2+sqrt2)/2$
${sin}^{2} t = \frac{2 + \sqrt{2}}{4}$ $sin^2 t = (2 + sqrt2)/4$
$sin t = \frac{\sqrt{2 + \sqrt{2}}}{2}$ $sin t = sqrt(2 + sqrt2)/2$ . Only the positive answer is accepted since the arc 67.5 is in Quadrant I.
$sin t = sin (67.5) = \frac{\sqrt{2 + \sqrt{2}}}{2}$ $sin t = sin (67.5) = sqrt(2 + sqrt2)/2$ .

Check by calculator>
sin (67.5) = 0.92
$\frac{\sqrt{2 + \sqrt{2}}}{2} = \frac{1.85}{2} = 0.92$ $sqrt(2 + sqrt2)/2 = 1.85/2 = 0. 92$ . OK

Science

Math

Humanities

... and beyond

How do you use the half-angle identity to find the exact value of sin67.5°?

1 Answer

Explanation:

Related questions

Impact of this question