How do you use half angle formula to find $sin 67.5$ $sin 67.5$ ?

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How do you use half angle formula to find $sin 67.5$ $sin 67.5$ ?

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Answer 1 · 2018-05-04T04:26:09

$\frac{\sqrt{2 + \sqrt{2}}}{2}$ $sqrt(2+sqrt2)/2$

Explanation:

${sin 67.5}^{\circ}$ $sin67.5^@$

The half angle formula for $sin (\frac{θ}{2}) = \pm \sqrt{\frac{1 - cos θ}{2}}$ $sin(theta/2) = +-sqrt((1-costheta)/2)$ .

The $\pm$ $+-$ sign depends on which quadrant your angle is in. Since ${67.5}^{\circ}$ $67.5^@$ is in the 1st quadrant and sine is positive in that quadrant, we know that it is positive.

${67.5}^{\circ} \cdot 2 = 135^{\circ}$ $67.5^@ * 2 = 135^@$ , so:
$sin (\frac{θ}{2}) = \sqrt{\frac{1 - {cos 135}^{\circ}}{2}}$ $sin(theta/2) = sqrt((1- cos135^@)/2)$

$quadquadquadquadquadquadquad=sqrt((1-(-sqrt2/2))/2)$

$quadquadquadquadquadquadquad=sqrt((2/2+sqrt2/2)/2)$

$quadquadquadquadquadquadquad=sqrt(((2+sqrt2)/2)/2)$

$quadquadquadquadquadquadquad=sqrt((2+sqrt2)/4)$

$quadquadquadquadquadquadquad=sqrt(2+sqrt2)/2$

Hope this helps!

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How do you use half angle formula to find $sin 67.5$ $sin 67.5$ ?

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