Help with simplest form sin(2pift - pi) + sin(2pift + pi)
Not sure what the f is doing in there, but recall that
sinA + sinB = 2sin((A+B)/2)cos((A-B)/2)
I think you got too many π's floating around in there, as well.
sin(x+π) = sin(x-π) = -sin(x)
It occurred to me this might be a physics problem:
sin(ωt-π)+sin(ωt+π)
= -2sin(ωt)
To simplify the expression sin(2πft - π) + sin(2πft + π), we can make use of the sum-to-product identities for trigonometric functions.
The sum-to-product identity for sine states that sin(A) + sin(B) = 2 * sin((A + B)/2) * cos((A - B)/2).
In our expression, we have sin(2πft - π) + sin(2πft + π). We can rewrite this as:
sin(2πft - π) + sin(2πft + π) = 2 * sin((2πft - π + 2πft + π)/2) * cos((2πft - π - 2πft - π)/2)
Simplifying the above expression, we get:
= 2 * sin(4πft/2) * cos(-2π/2)
= 2 * sin(2πft) * cos(-π)
= 2 * sin(2πft) * (-1)
= -2 * sin(2πft)
So, sin(2πft - π) + sin(2πft + π) simplifies to -2 * sin(2πft).
To simplify the expression sin(2pift - pi) + sin(2pift + pi), we can use the angle addition formula for sine. The formula states that sin(a + b) = sin(a)cos(b) + cos(a)sin(b).
In this case, a = 2pift and b = -pi. Applying the formula, we have:
sin(2pift - pi) + sin(2pift + pi)
= sin(2pift)cos(-pi) + cos(2pift)sin(-pi) + sin(2pift)cos(pi) + cos(2pift)sin(pi)
Since cos(-pi) = cos(pi) = -1 and sin(-pi) = sin(pi) = 0, we can simplify further:
sin(2pift - pi) + sin(2pift + pi)
= sin(2pift)(-1) + cos(2pift)(0) + sin(2pift)(-1) + cos(2pift)(0)
= -sin(2pift) - sin(2pift)
= -2sin(2pift)
Therefore, the simplified form of sin(2pift - pi) + sin(2pift + pi) is -2sin(2pift).