Question

Prove that $\sin {30}^o\cdot \cos {60}^o+\cos {30}^o\cdot \sin {60}^o=\sin {90}^o$

Answer 1

We know that the values of the trignometric functions $\sin {60}^0=\frac{\sqrt{3}}{2}$, $\cos {30}^0=\frac{\sqrt{3}}{2}$, $\sin {30}^0=\frac{1}{2}$,$\cos {60}^0=\frac{1}{2}$ and $\sin {90}^0=1$.

Let usfirstfind the value of left hand side (LHS) of the given identity $\sin {30}^0\cos {60}^0+\cos {30}^0\sin {60}^0=\sin {90}^0$ as shown below:

$\sin {30}^0\cos {60}^0+\cos {30}^0\sin {60}^0=\left(\frac{1}{2}\times \frac{1}{2}\right)+\left(\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}\right)=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1...........\left(1\right)$

Now, we find the value of right hand side RHS) of the given identity $\sin {30}^0\cos {60}^0+\cos {30}^0\sin {60}^0=\sin {90}^0$ as follows:

$\sin {90}^0=1........\left(2\right)$

From equations 1 and 2, we get that LHS=RHS and

Hence, $\sin {30}^0\cos {60}^0+\cos {30}^0\sin {60}^0=\sin {90}^0$.