Question

If $B={15}^o$, prove that $4\sin 2B\cdot \cos 4B\cdot \sin 6B=1$

Answer 1

In the given identity $4\sin 2B\cos 4B\sin 6B$, substitute $B={15}^0$ as shown below:

$\sin 2B\cos 4B\sin 6B\Rightarrow 4\sin (2\times {15}^0)\cos (4\times {15}^0)\sin (6\times {15}^0)\Rightarrow 4\sin {30}^0\cos {60}^0\sin {90}^0$

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

Let usfind the value of the left hand side (LHS) that is $4\sin {30}^0\cos {60}^0\sin {90}^0$ as shown below:

$4\sin {30}^0\cos {60}^0\sin {90}^0=(4\times \frac{1}{2})\times \frac{1}{2}\times 1=2\times \frac{1}{2}=1=RHS$

Since LHS=RHS,

Hence, $4\sin 2B\cos 4B\sin 6B=1$ when $B={15}^0$.