If the value of 3sin3B+2cos(2B+5∘)2cos3B−sin(2B−10∘); when B= 20∘ is 3√m+2√2, then find the value of m.
3sin3B+2cos(2B+5∘)2cos3B−sin(2B−10∘); when B = 20∘
when, B=20∘
3sin60∘+2cos(45∘)2cos60∘−sin(30∘)
=3(√32)+2(1√2)2(12)−(12)
=(3√32)+(2√22)(22)−(12)
=3√3+2√2