since sin2A=2sinAcosA, what is sin1/2A?
isn't it just equals to sin1/2Acos1/2A?
Sin2a = 2sinAcosA If we apply inverse sin to both sides of the equation: 2A = sin^-1(2sinAcosA) sin1/2A sin1/sin^-1(2sinAcosA)
why must we inverse both sides?
When you inverse both sides of the equation you make the angle (2a) by itself. This is necisary in solving sin1/2a. In the denominator 2a is by itself, and not the sin,
what is the ans for sin1/2A then?
It can only be solved in terms of variables; seeing is though it cannot be solved in any other methods.
sin1/2a = sin1/sin^-1(2sinAcosA)
actually i got this result, sin2A=2sinAcosA so i was thinking, can't we replace A with 1/2A to get sin1/2A? ie sin1/2A = 2sin1/2Acos1/2A
No, you cannot replace A with 1/2A..... How can you replace A with 1/2A. They are different numbers
then how did this step come about? from 4sin1/2xcos1/2x =2sinx it's a solution from my math book
I don't know how your maths book achieved any answer. I don't know where the x came from either. The equation can only be simplified.
Or solved in terms of variables. This is what I have done
Is it sin A/2 or 1/(2A)?
the second one
U can write 1/2A = 1/4A + 1/4A and use double angle if u want...
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