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Simplify (cos5theta-cos theta)/ (sin5theta-sin theta) @mathmate
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= cos5 cos - sin5 sin =..
@eric_d Are the "5"s exponents or multiplicators?
\[\cos 5 \theta ...\]
shld be multiplication
the given answer is -tan 3 theta
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@eric do u know tht cosA - cosB = 2sin[(A+B)/2]*sin[(B-A)/2]
I got to go soon, but hope these identities help you: \(cos(5t)=5*cos(t)sin^4(t)-10cos^3(t)sin^2(t)+cos^5(t)\) \(sin(5t)=sin^5(t)-10cos^2(t)sin^3(t)+5cos^4(t)sin(t)\) and \(\large -tan(3t)=\frac{tan^3(t)-3tan(t)}{1-3tan^2(t)}\)
Yes, I do. This set of identities is very useful.
Will talk later!
sin A - sinB = 2cos[(A+B)/2]*sin[(A-B)/2]
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Thanks.. @mathmate
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