What is the value of Sin(75)?

Question: What is the value of Sin(75)?

Solution :

We shall write \[75^0=45^0+30^0 \] And apply Sin(A+B) formula \[Sin(A+B) =Sin(A)Cos(B)+Cos(A)Sin(A)\] Therefore \[Sin(75^0) = Sin(45^0+30^0) =Sin(45^0)Cos(30^0)+Cos(45^0)Sin(30^0) \] Now we shall substitute following values into above formulas \[Sin(45^0) =1/√2 , Cos(30^0) = √3/2 , Cos(45^0) = 1/√2 , Sin(30^0) =1/2 \] That is \[Sin\left({75}^0\ \right)=\frac{1}{\sqrt2}\frac{\sqrt3}{2}+\frac{1}{\sqrt2}\frac{1}{2}\]\[ Sin\left({15}^0 \right)=\frac{1}{2\sqrt2}\left(\sqrt3+1\right)\]

Thus answer of the question What is the value of Sin(75)? is

\[ \frac{1}{2\sqrt2}\left(\sqrt3+1\right)\]