You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
Trigonometric Identities
- Thread starter Goistein
- Start date
Goistein said:How do you get from multiplictions of sines/cosines to additions? Thanks in advance!
How about posting an example of a problem you're having trouble with? That way, we will have something specific to work with. As it is, your question is pretty vague and that is probably why you haven't received a response so far.
Hello, Goistein!
I'm still not sure what you're looking for, but . . .
There are four Product-to-Sum formulas:
. . \(\displaystyle \sin(A)\cdot\sin(B) \;=\;\frac{1}{2}\left[\cos(A-B)\,-\,\cos(A+B)\right]\)
. . \(\displaystyle \sin(A)\cdot\cos(B)\;=\;\frac{1}{2}\left[\sin(A-B)\,+\,\sin(A+B)\right]\)
. . \(\displaystyle \cos(A)\cdot\cos(B)\;=\;\frac{1}{2}\left[\cos(A-B)\,+\,\cos(A+B)\right]\)
. . \(\displaystyle \cos(A)\cdot\sin(B)\;=\;\frac{1}{2}\left[\sin(A-B)\,-\,\sin(A-B)\right]\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
I'll derive one of them . . .
We have: \(\displaystyle \:\begin{array}{cc}\sin(A+B)\:=\:\sin(A)\cos(B)\,+\,\sin(B)\cos(A) & \;\;[1]\\ \sin(A-B)\:=\:\sin(A)\cos(B)\,-\,\sin(B)\cos(A) & \;\;[2]\end{array}\)
Add [1] and [2]: \(\displaystyle \:\sin(A+B)\,+\,\sin(A-B)\;=\;2\sin(A)\cos(B)\)
Therefore: \(\displaystyle \:\sin(A)\cos(B)\;=\;\frac{1}{2}\left[\sin(A+B)\,+\,\sin(A-B)\right]\)
I'm still not sure what you're looking for, but . . .
How do you get from multiplictions of sines/cosines to additions?
There are four Product-to-Sum formulas:
. . \(\displaystyle \sin(A)\cdot\sin(B) \;=\;\frac{1}{2}\left[\cos(A-B)\,-\,\cos(A+B)\right]\)
. . \(\displaystyle \sin(A)\cdot\cos(B)\;=\;\frac{1}{2}\left[\sin(A-B)\,+\,\sin(A+B)\right]\)
. . \(\displaystyle \cos(A)\cdot\cos(B)\;=\;\frac{1}{2}\left[\cos(A-B)\,+\,\cos(A+B)\right]\)
. . \(\displaystyle \cos(A)\cdot\sin(B)\;=\;\frac{1}{2}\left[\sin(A-B)\,-\,\sin(A-B)\right]\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
I'll derive one of them . . .
We have: \(\displaystyle \:\begin{array}{cc}\sin(A+B)\:=\:\sin(A)\cos(B)\,+\,\sin(B)\cos(A) & \;\;[1]\\ \sin(A-B)\:=\:\sin(A)\cos(B)\,-\,\sin(B)\cos(A) & \;\;[2]\end{array}\)
Add [1] and [2]: \(\displaystyle \:\sin(A+B)\,+\,\sin(A-B)\;=\;2\sin(A)\cos(B)\)
Therefore: \(\displaystyle \:\sin(A)\cos(B)\;=\;\frac{1}{2}\left[\sin(A+B)\,+\,\sin(A-B)\right]\)