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trig identities
- Thread starter feeble
- Start date
D
Deleted member 4993
Guest
feeble said:sin(x+y)sin(x-y)=cos^2y-cos^2x I'm totally lost help....
Hint:
\(\displaystyle sin(A) * sin(B) \ = \ \frac{cos(A-B) - cos(A+B)}{2}\)
Please share your work with us, indicating exactly where you are lost - so that we may know where to begin to help you.
Hello, feeble!
Did you try anything?
\(\displaystyle \sin(x+y)\sin(x-y) \;=\;(\sin x\cos y + \cos x\sin y)(\sin x\cos y - \cos x\sin y)\)
. . . . . . . . . . . . . . \(\displaystyle =\;\sin^2\!x\cos^2\!y - \cos^2\!x\sin^2\!y\)
. . . . . . . . . . . . . . \(\displaystyle =\;(1-\cos^2\!x)\cos^2\!y - \cos^2\!x(1-\cos^2\!y)\)
. . . . . . . . . . . . . . \(\displaystyle =\;\cos^2\!y - \cos^2\!x\cos^2\!y - \cos^2\!x + \cos^2\!x\cos^2\!y\)
. . . . . . . . . . . . . . \(\displaystyle =\;\cos^2\!y - \cos^2\!x\)
Did you try anything?
\(\displaystyle \sin(x+y)\sin(x-y) \:=\:\cos^2\!y - \cos^2\!x\)
\(\displaystyle \sin(x+y)\sin(x-y) \;=\;(\sin x\cos y + \cos x\sin y)(\sin x\cos y - \cos x\sin y)\)
. . . . . . . . . . . . . . \(\displaystyle =\;\sin^2\!x\cos^2\!y - \cos^2\!x\sin^2\!y\)
. . . . . . . . . . . . . . \(\displaystyle =\;(1-\cos^2\!x)\cos^2\!y - \cos^2\!x(1-\cos^2\!y)\)
. . . . . . . . . . . . . . \(\displaystyle =\;\cos^2\!y - \cos^2\!x\cos^2\!y - \cos^2\!x + \cos^2\!x\cos^2\!y\)
. . . . . . . . . . . . . . \(\displaystyle =\;\cos^2\!y - \cos^2\!x\)
D
Deleted member 4993
Guest
Another way:
\(\displaystyle sin(x+y)\cdot sin(x-y) \\)
\(\displaystyle = \ \frac{[cos(x+y-x+y) + cos(x+y+x-y)]}{2} \\)
\(\displaystyle = \ \frac{cos(2y) - cos(2x)}{2} \\)
\(\displaystyle = \ \frac{2cos^2(y) - 1 - 2cos^(x) + 1}{2} \\)
\(\displaystyle = \ cos^2(y) - cos^2(x)\)
\(\displaystyle sin(x+y)\cdot sin(x-y) \\)
\(\displaystyle = \ \frac{[cos(x+y-x+y) + cos(x+y+x-y)]}{2} \\)
\(\displaystyle = \ \frac{cos(2y) - cos(2x)}{2} \\)
\(\displaystyle = \ \frac{2cos^2(y) - 1 - 2cos^(x) + 1}{2} \\)
\(\displaystyle = \ cos^2(y) - cos^2(x)\)