Question #13c72

1 Answer
Steve M
Feb 4, 2018

We seek to prove that:

# sin(npi + theta) -= (-1)^n(sintheta) #

We can use the sine sum of angle identity:

# sin(A+B) -= sinAcosB+cosAsinB #

Applying this to the LHS, we get:

# sin(npi + theta) -= sin(npi)costheta+cos(npi)sintheta #

Now, if we consider the graphs of the trigonometric functions:

http://community.boredofstudies.org

We can see that:

# sin(npi) = 0 #
# cos(npi) = { (-1, n " odd"), (1, n " even") :} = (-1)^n #

Hence we have the given result:

# sin(npi + theta) -= 0 + (-1)^nsintheta #
# " " -= (-1)^nsintheta \ \ \ # QED

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