does sin2(x)=2 sinx cosx?
Really???
if by sin2(x) you mean sin^2(x) then
not always
clearly, if x=0, sin^x(x) = sin(2x) = 0
Otherwise, you have
sin^2(x) = 2 sinx cosx
sinx = 2 cosx
tanx = 2
x = arctan(2)
What I'm asking is did you solve sin2[sin^-1(pi/6)] and sin[2sin^-1(pi/6)] in the same way?
To verify if sin²(x) is equal to 2sin(x)cos(x), we can use the trigonometric identity for double angle, which states that sin²(x) = 1/2(1 - cos(2x)).
Let's substitute this identity into the equation:
sin²(x) = 1/2(1 - cos(2x))
Now, let's simplify the right-hand side of the equation:
sin²(x) = 1/2 - 1/2cos(2x)
Here, we can clearly see that sin²(x) does not equal 2sin(x)cos(x). The correct expression for sin²(x) is given by the trigonometric identity mentioned earlier, sin²(x) = 1/2(1 - cos(2x)).
So, the statement sin²(x) = 2sin(x)cos(x) is not true.
no. You asked if they were equal.
I showed you that they are, only in two special cases.
Your followup question is in no way similar.
sin^-1(pi/6) makes no sense at all. I mean, it's possible, since pi/6 < 1 so there is some angle such that sin(x) = pi/6.