How do I find the value of sin 105?

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How do I find the value of sin 105?

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Answer 1 · 2015-09-24T17:47:22

It is ${sin 105}^{o} = \frac{1}{4} \cdot (\sqrt{6} + \sqrt{2})$ $sin105^o=1/4*(sqrt6+sqrt2)$

Explanation:

We have that

$sin (105) = sin (60 + 45)$ $sin(105)=sin(60+45)$

because

$sin (a + b) = sin a \cdot cos b + cos a \cdot sin b$ $sin(a+b)=sina*cosb+cosa*sinb$

hence, we have

$sin 105 = sin (60 + 45) = sin 60 \cdot cos 45 + cos 60 \cdot sin 45 = = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} + \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = \frac{1}{4} (\sqrt{6} + \sqrt{2})$ $sin105=sin(60+45)=sin60*cos45+cos60*sin45= =sqrt3/2*sqrt2/2+1/2*sqrt2/2=1/4(sqrt6+sqrt2)$

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How do I find the value of sin 105?

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