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trig identity help
- Thread starter nfs
- Start date
Hello, nfs!
\(\displaystyle \text{Factor: }\;(\sin^2\!x - \cos^2\!x)\underbrace{(\sin^2\!x + \cos^2\!x)}_{\text{This is 1}}\)
. . . . . . \(\displaystyle =\quad\sin^2\!x - \cos^2\!x\)
. . . . . . \(\displaystyle =\; \sin^2\!x - \overbrace{(1 - \sin^2\!x)}\)
. . . . . . \(\displaystyle =\;\sin^2\!x - 1 + \sin^2\!x\)
. . . . . . \(\displaystyle =\; 2\sin^2\!x-1\)
\(\displaystyle \text{Verify: }\;\sin^4\!x - \cos^4\!x \;=\;2\sin^2\!x - 1\)
\(\displaystyle \text{Factor: }\;(\sin^2\!x - \cos^2\!x)\underbrace{(\sin^2\!x + \cos^2\!x)}_{\text{This is 1}}\)
. . . . . . \(\displaystyle =\quad\sin^2\!x - \cos^2\!x\)
. . . . . . \(\displaystyle =\; \sin^2\!x - \overbrace{(1 - \sin^2\!x)}\)
. . . . . . \(\displaystyle =\;\sin^2\!x - 1 + \sin^2\!x\)
. . . . . . \(\displaystyle =\; 2\sin^2\!x-1\)
D
Deleted member 4993
Guest
nfs said:thanks but how do you get from the second last step to the last step?
Get pencil and paper and write it out - you'll see that it is very conspicuous.
nfs said:thanks but how do you get from the second last step to the last step?
oh i understand it now thanks guys