it's pi instead of 1
sin(x+pi) = -sin(x)
solve sin(x+1) = -sinx using the compound angle formula
I know it's sin(a+b) = sin(a)cos(b) + cos(a)sin(b) but idk what should I do next
3 answers
sin(x+1) = -sinx
sinxcos1 + cosxsin1 = -sinx
divide each term by sinx
cos1 + cosxsin1/sinx = -1
cotx sin1 = -1-cos1
cotx = (-1-cos1)/sin1
tanx = -sin1/(1 + cos1) = -.5463..... so x is in quad II or IV
x = -.5 , by my calculator
but the period of tanx is π
so we have
x = -.5, -.5±π, -.5±2π , etc
for 0 ≤ x ≤ 2π we get x = appr 2.6416, 5.7832
sinxcos1 + cosxsin1 = -sinx
divide each term by sinx
cos1 + cosxsin1/sinx = -1
cotx sin1 = -1-cos1
cotx = (-1-cos1)/sin1
tanx = -sin1/(1 + cos1) = -.5463..... so x is in quad II or IV
x = -.5 , by my calculator
but the period of tanx is π
so we have
x = -.5, -.5±π, -.5±2π , etc
for 0 ≤ x ≤ 2π we get x = appr 2.6416, 5.7832
Oh well, now it becomes really easy
your expansion is simply
sinxcosπ + cosxsinπ = -sinx
we know cosπ = -1 and sinπ = 0
so we get
-sinx + 0 = -sinx
0 = 0
so the equation is just an identity, and it is true for all values of x
your expansion is simply
sinxcosπ + cosxsinπ = -sinx
we know cosπ = -1 and sinπ = 0
so we get
-sinx + 0 = -sinx
0 = 0
so the equation is just an identity, and it is true for all values of x