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Please help i beg i never get these right

Please help i beg i never get these right-example-1
User Arash Rabiee
by
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1 Answer

16 votes
16 votes

Answer:


\sin(105) = (\sqrt 2 + \sqrt 6)/(4)

Explanation:

Given


\sin(105^o)

Required

Solve

Using sine rule, we have:


\sin(A + B) = \sin(A)\cos(B) + \sin(B)\cos(A)

This gives:


\sin(105^o) = \sin(60 + 45)

So, we have:


\sin(60 + 45) = \sin(60)\cos(45) + \sin(45)\cos(60)

In radical forms, we have:


\sin(60 + 45) = (\sqrt 3)/(2) * (\sqrt 2)/(2) + (\sqrt 2)/(2) * (1)/(2)


\sin(60 + 45) = (\sqrt 6)/(4) + (\sqrt 2)/(4)

Take LCM


\sin(60 + 45) = (\sqrt 6 + \sqrt 2)/(4)

Rewrite as:


\sin(60 + 45) = (\sqrt 2 + \sqrt 6)/(4)

Hence:


\sin(105) = (\sqrt 2 + \sqrt 6)/(4)

User JaredMcAteer
by
2.6k points