How do you use the half angle identity to find sin 105?

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How do you use the half angle identity to find sin 105?

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Answer 1 · 2018-03-20T02:07:09

${sin 105}^{\circ} = \frac{\sqrt{2 + \sqrt{3}}}{2}$ $sin105^@=sqrt(2+sqrt3)/2$

Explanation:

We need to use the half angle formula:

$sin (\frac{θ}{2}) = \pm \sqrt{\frac{1 - cos θ}{2}}$ $sin(theta/2)=+-sqrt((1-costheta)/2)$

In this case, we want to find $sin (105^{\circ})$ $sin(105^@)$ , so that's what we want $sin (\frac{θ}{2})$ $sin(theta/2)$ to equal. To find out what our $θ$ $theta$ is, set these to equal to each other:

$sin (105^{\circ}) = sin (\frac{θ}{2})$ $sin(105^@)=sin(theta/2)$

$105^{\circ} = \frac{θ}{2}$ $105^@=theta/2$

$210^{\circ} = θ$ $210^@=theta$

This is our $θ$ $theta$ . Now, we can use the half angle formula:

$= sin (105^{\circ})$ $color(white)=sin(105^@)$

$= sin (\frac{210^{\circ}}{2})$ $=sin(210^@/2)$

$= \pm \sqrt{\frac{1 - cos (210^{\circ})}{2}}$ $=+-sqrt((1-cos(210^@))/2)$

$= \pm \sqrt{\frac{1 - (- \frac{\sqrt{3}}{2})}{2}}$ $=+-sqrt((1-(-sqrt3/2))/2)$

$= \pm \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}}$ $=+-sqrt((1+sqrt3/2)/2)$

$= \pm \sqrt{\frac{2 + \sqrt{3}}{4}}$ $=+-sqrt((2+sqrt3)/4)$

$= \pm \frac{\sqrt{2 + \sqrt{3}}}{\sqrt{4}}$ $=+-sqrt(2+sqrt3)/sqrt4$

$= \pm \frac{\sqrt{2 + \sqrt{3}}}{2}$ $=+-sqrt(2+sqrt3)/2$

Since $105^{\circ}$ $105^@$ is in quadrant II, we know that our answer will be positive that angle is above the $x$ $x$ -axis (and we are taking the sine). Therefore:

${sin 105}^{\circ} = \frac{\sqrt{2 + \sqrt{3}}}{2}$ $sin105^@=sqrt(2+sqrt3)/2$

We can check our answer using a calculator (be sure it is in degrees mode):

![https://www.desmos.com/calculator]( useruploads.socratic.org )

Answer 2 · 2018-03-20T02:31:55

$sin (105) = = \pm (\frac{1}{2}) \sqrt{2 + \sqrt{3}}$ $color(blue)(sin (105) = = +-(1/2) sqrt(2 + sqrt3)$

Explanation:

![ www2.clarku.edu )

Given $\frac{θ}{2} = 105^{\circ} = \frac{7 π}{12}$ $theta / 2 = 105^@ = (7pi)/12$

$θ = \frac{7 π}{6}$ $theta = (7pi)/6$

$sin (\frac{θ}{2}) = \pm \sqrt{\frac{1 - cos θ}{2}}$ $sin (theta/2) = +- sqrt((1 - cos theta)/2)$

$sin (\frac{7 π}{12}) = \pm \sqrt{\frac{1 - cos ((\frac{7 π}{12}) \cdot 2)}{2}}$ $sin ((7pi)/12) = +- sqrt((1-cos (((7pi)/12)*2))/2)$

$sin (\frac{7 π}{12}) = \pm \sqrt{\frac{1 - cos (\frac{7 π}{6})}{2}}$ $sin ((7pi)/12) = +- sqrt((1 - cos ((7pi)/6))/2)$

But $cos (\frac{7 π}{6}) = cos (π + \frac{π}{6}) = - cos (\frac{π}{6})$ $cos ((7pi)/6) = cos (pi + pi/6) = - cos (pi/6)$

$:. sin ((7pi)/12) = +- sqrt((1 + cos(pi/6))/2)$

$=> +- sqrt((1 + sqrt(3)/2)/2)$

$=> +- sqrt((2 + sqrt3)/4) = +-(1/2) sqrt(2 + sqrt3)$

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How do you use the half angle identity to find sin 105?

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