Show that sin(x+pi)=-sinx.
So far, I used the sum formula for sin which is sin(a+b)=sin a cos b+cos a sin b.
sin(x+pi)=sin x cos pi+cos x sin pi
I think I am supposed to do this next, but I am not sure.
sin(x+pi)=sin x cos x+sin pi cos pi
If that is right then I am not sure what to do from there. Can someone please help?
in(x+pi)=sin x cos pi+cos x sin pi but cosPI=-1, sin PI=0
sin(x+PI)= -sinx
Thanks bobpursley!!!
Can you please help me with one more?
Which are simplified forms of the expression tan^2T cos2T? Select all that apply. (2 answers)
a. 2sin^2T-tan^2T
b. tan^2T-2cos^2T
c. 2sin^3T/cosT
d. cos2T/cot^2T
e. 2cos^2T-tan^2T
(sin^2T/cos^2T)(2 cos^2 T - 1)
= 2 sin^2 T - tan^2 T
To prove that sin(x+π) = -sin(x), we can use the sum formula for sine, as you have correctly done.
Starting from where you left off:
sin(x+π) = sin(x)cos(π) + cos(x)sin(π)
Now, we need to simplify the expression further. Let's break down the values of sine and cosine for π:
sin(π) = 0
cos(π) = -1
Substituting these values, we get:
sin(x+π) = sin(x)(-1) + cos(x)(0)
Since sin(π) = 0, it disappears from the equation. Additionally, multiplying any number by 0 gives a result of 0. Therefore, we have:
sin(x+π) = -sin(x) + 0
Finally, removing the 0 term, we arrive at the desired conclusion:
sin(x+π) = -sin(x)
Hence, sin(x+π) is indeed equal to -sin(x).