Show that sin(x+pi)=-sinx. Show all of your work.
use the expansion equation
sin(A+B) = sinAcosB + cosAsinB
sin(x + π) = sinxcosπ + cosxsinπ
= sinx(-1) + cosx(0)
= -sinx
sin(xπi) = sinx cosπ + cosx sinπ
Thank you Reiny and oobleck!
Okay, let's dive into the world of trigonometry!
To prove that sin(x+π) = -sin(x), we can use some handy trigonometric identities. For clarity, let's start from the left-hand side (LHS) of the equation.
LHS: sin(x + π)
We know that π is the same as 180 degrees, so we can rewrite the equation as:
sin(x + 180°)
Now, let's utilize the sum formula for sine:
sin(A + B) = sin(A)cos(B) + cos(A)sin(B)
In our case, A = x and B = 180°. Plug in these values:
sin(x + 180°) = sin(x)cos(180°) + cos(x)sin(180°)
Next, we need to determine the values of cos(180°) and sin(180°). Luckily, we have some trigonometric values handy!
cos(180°) = -1
sin(180°) = 0
Substituting these values into the equation, we get:
sin(x + 180°) = sin(x)(-1) + cos(x)(0)
Simplifying further:
sin(x + 180°) = -sin(x) + 0
Final step:
sin(x + 180°) = -sin(x)
And voila! We have proven that sin(x + π) = -sin(x) using some trigonometric identities and calculations.
Remember, math can be funny sometimes, but it always adds up!
To prove that sin(x + π) = -sin(x), we can use the properties of the sine function and the angle addition formula.
Let's start with the angle addition formula for the sine function:
sin(a + b) = sin(a)⋅cos(b) + cos(a)⋅sin(b)
In our case, a = x and b = π:
sin(x + π) = sin(x)⋅cos(π) + cos(x)⋅sin(π)
Now, let's apply the values of cos(π) and sin(π):
cos(π) = -1 and sin(π) = 0
Substituting these values:
sin(x + π) = sin(x)⋅(-1) + cos(x)⋅0
Since cos(π) = -1 and sin(π) = 0, we can simplify the equation further:
sin(x + π) = -sin(x) + 0
simplifying the equation:
sin(x + π) = -sin(x)
Therefore, we have shown that sin(x + π) is equal to -sin(x).