Typical sum and difference formulas usually go in the order of :
Cos(A+B) or Cos(A-B)
for this variation of the formula I am asked to solve for Cos(B-A).
What I attempted doing was switching the original formula around like so...
Cos(B-A) = Sin(A)*Sin(B) + Cos(a)*Cos(B)
But that yielded an incorrect answer.
Another attempt I tired was switching the variables instead of the trig functions but that was also incorrect. Can someone tell me what I have done wrong?
$\endgroup$ 12 Answers
$\begingroup$Start with $\cos(A-B) = \cos A\cos B + \sin A \sin B$
To work out $\cos (B-A)$, you can proceed in two logical ways. By the first way, you can just transpose (swap) $A$ and $B$ in every step. To get:
$\cos (B-A) = \cos B \cos A + \sin B \sin A$
Of course, because of the commutative rule of multiplication, this works out to give you exactly the same expression as the original, i.e. you can rearrange this to:
$\cos (B-A) = \cos A \cos B + \sin A \sin B$
By the second way, you can use the property that the cosine function is even. This means that $\cos (-x) = \cos x$ for any $x$.
Since $(B-A) = -(A-B)$,
$\cos (B-A) = \cos(A-B)= \cos A \cos B + \sin A \sin B$
so you end up with the same result.
Now the expression you wrote down: $\sin A \sin B + \cos A \cos B$ looks like you just swapped the order of the trig product terms around. That doesn't look like sound logic. However, the end result is still (coincidentally, probably) correct because addition is commutative. So your answer should still have been correct.
If you did get an incorrect answer, you should clarify your question with more details.
$\endgroup$ $\begingroup$$$\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)$$ $$\cos(A-B)=\cos(A)\cos(B)+\sin(A)\sin(B)$$
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