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Trig identitities - make the left side look like the right side without changing the right side
( cos^2(x) - sin^2(x) ) / (sinx + cosx) = cosx - sinx
5 Answers
- DWReadLv 78 years agoFavorite Answer
cos²θ - sin²θ is the difference of squares, so cos²θ - sin²θ = (cosθ + sinθ)(cosθ - sinθ)
(cos²θ - sin²θ)/(sinθ + cosθ) = (cosθ + sinθ)(cosθ - sinθ)/(sinθ + cosθ)
= cosθ - sinθ
- 8 years ago
cos^2(x)-sin^2(x) = (cos(x)+sin(x))(cos(x)-sin(x))
If that is above (sin(x)+cos(x)) then you can cancel out the cos(x)+sin(x) because they're the same thing
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- JOS JLv 78 years ago
(Cos[x]^2 - Sin[x]^2)/(Sin[x] + Cos[x]) = Cos[x] - Sin[x]
Cos[2 x]/(Cos[x] + Sin[x]) = Cos[x] - Sin[x]
Cos[x] - Sin[x] = Cos[x] - Sin[x]
