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
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
it's pi instead of 1
sin(x+pi) = -sin(x)
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
To solve the equation sin(x+1) = -sin(x) using the compound angle formula, we can rewrite the equation using the formula sin(a+b) = sin(a)cos(b) + cos(a)sin(b).
Let's apply the formula to the equation sin(x+1) = -sin(x):
sin(x)cos(1) + cos(x)sin(1) = -sin(x)
Now, we can rearrange the equation:
sin(x)cos(1) + sin(x)sin(1) + sin(x) = 0
Next, combine the like terms:
sin(x)(cos(1) + sin(1) + 1) = 0
To solve this equation, we have two possibilities for sin(x):
1. sin(x) = 0
In this case, x can take any value where sin(x) equals zero, such as x = 0, π, 2π, etc.
2. cos(1) + sin(1) + 1 = 0
To find the value of x for this case, we need to solve the equation:
cos(1) + sin(1) + 1 = 0
Unfortunately, there is no algebraic solution for this equation. We can use numerical methods or a calculator to find an approximate solution for x.
By using the compound angle formula, we have rewritten the equation sin(x+1) = -sin(x) into sin(x)(cos(1) + sin(1) + 1) = 0. From there, we can find the values of x that satisfy the equation.