Question

plzz solve this....​
as far as u all can

Answer 1

Approach:

The idea was to use the identity $\sin{(a+b)} \sin{(a-b)} = sin^2{a}-sin^2{b}$ and then simplify to get sin(3x). This problem is pretty simple provided you remember the above result.

Don't worry if you didn't know about it. It's pretty easy to see why that must be true. Try proving that as an exercise :)

Step-by-step Proof:

$LHS\\=4\sin{(x)} \sin{(\frac{\pi}{3} - x)}\sin{(\frac{\pi}{3} +x)}\\= 4 \sin{x} \times (sin^2{\frac{\pi}{3}}-\sin^2{x})\\(\because \sin{(a+b)} \sin{(a-b)} = sin^2{a}-sin^2{b}  \forall a, b \in R)\\= 4\sin{x} \times (\frac{3}{4}-\sin^2{x})\\\text{(Expand this expression)}\\=(\frac{4}{4})3 \sin{x}-4\sin^3{x}\\\text{(Cancel the 4 from the numerator and denominator)}\\=3\sin{x}-4sin^3{x}\\=\sin{3x}\\=RHS$

Hence, proved

I hope that helped.