Prove that, in general, sin(alpha + beta) does not equal sin alpha + sin beta
all you need is a single couterexample to show it DOES NOT work
so pick 30 and 60
to show sin(30+60) is not equal to sin30 + sin 60
is sin90 = sin30 + sin60 ??
To prove that, in general, sin(alpha + beta) does not equal sin alpha + sin beta, we need to show a counterexample where the equation does not hold true.
Let's take alpha = 0 and beta = 90 degrees (π/2 radians).
Using the sum formula for sine, we have:
sin(alpha + beta) = sin(0 + π/2) = sin(π/2) = 1
On the other hand,
sin alpha + sin beta = sin(0) + sin(π/2) = 0 + 1 = 1
In this case, sin(alpha + beta) equals sin alpha + sin beta. Therefore, this is not a valid counterexample.
To further prove the claim, we need to show that the equation sin(alpha + beta) = sin alpha + sin beta does not hold for any other values of alpha and beta.
We can use a graphical approach to demonstrate this. Using trigonometric identities, we can rewrite the equation sin(alpha + beta) = sin alpha + sin beta as:
sin alpha * cos beta + cos alpha * sin beta = sin alpha + sin beta
Rearranging the equation, we get:
sin alpha * (1 - cos beta) - cos alpha * sin beta = 0
This equation represents the difference between two vectors in the xy-plane, each with components sin alpha and cos alpha. For the equation to hold true, the vectors must be collinear, which means one vector is a scalar multiple of the other.
However, since sin and cos functions have different characteristics, their vector representations cannot be collinear. Therefore, the equation sin(alpha + beta) = sin alpha + sin beta does not hold in general.
In conclusion, using a counterexample and a graphical approach, we have proven that sin(alpha + beta) does not equal sin alpha + sin beta in general.