I am stuck
- Thread starter Loki123
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BigBeachBanana
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Factor this first.
D
Deleted member 4993
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I would first factorize the given expression for 'y' and simplify to get to a compact expression. You would not need to differentiate.
How??
skeeter
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[math]y = \dfrac{\sin(4x)}{4}[/math]
Factor, simplify, and then do u-substitution to avoid mechanical errors.
[math]f(x) = y = \sin(x) \cos^3(x) - \sin^3(x) \cos(x) = \\ \sin(x) \cos(x) \{\cos^2(x) - \sin^2 (x) \} =\\ u(\cos^2 (x) - \{1 - \cos^2 (x) \} ) =\\ u(2 \cos^2 (x) - 1) = uv \implies \\ y’ = uv’ + u’v.[/math]
That was not too hard.
[math] u = \sin (x) \cos (x) \implies\\ u’ = \sin (x) \{ - \sin (x)\} + \cos (x) \cos (x) = \cos^2 (x) - \sin^2 (x) = 2 \cos^2 (x) - 1.[/math]
Basic, but you may notice that [imath]u’ = v[/imath]. Weird but very interesting.
[math]v = \cos^2(x) - 1 \implies v’ = 2 \cos (x) \{- \sin(x) \} = -2u.[/math]
Weirder and weirder (unless I screwed up somewhere).
[math]\therefore y’ = uv’ + u’v = - 2u^2 + v^2.\\ \therefore y’ = 0 \implies v^2 = 2u^2.[/math]
Check all that. If it is correct, keep on trucking by reducing everything to cosines. I suspect you will need another substitution at the end.
[math]f(x) = y = \sin(x) \cos^3(x) - \sin^3(x) \cos(x) = \\ \sin(x) \cos(x) \{\cos^2(x) - \sin^2 (x) \} =\\ u(\cos^2 (x) - \{1 - \cos^2 (x) \} ) =\\ u(2 \cos^2 (x) - 1) = uv \implies \\ y’ = uv’ + u’v.[/math]
That was not too hard.
[math] u = \sin (x) \cos (x) \implies\\ u’ = \sin (x) \{ - \sin (x)\} + \cos (x) \cos (x) = \cos^2 (x) - \sin^2 (x) = 2 \cos^2 (x) - 1.[/math]
Basic, but you may notice that [imath]u’ = v[/imath]. Weird but very interesting.
[math]v = \cos^2(x) - 1 \implies v’ = 2 \cos (x) \{- \sin(x) \} = -2u.[/math]
Weirder and weirder (unless I screwed up somewhere).
[math]\therefore y’ = uv’ + u’v = - 2u^2 + v^2.\\ \therefore y’ = 0 \implies v^2 = 2u^2.[/math]
Check all that. If it is correct, keep on trucking by reducing everything to cosines. I suspect you will need another substitution at the end.
D
Deleted member 4993
Guest
y = sin(x)*cos(x)*{cos2(x)−sin2(x)}
= (1/2) * sin(2x) * cos(2x)
= (1/4) sin(4x) ............................................................. (response #3 & #4)
CONTINUE......
= (1/2) * sin(2x) * cos(2x)
= (1/4) sin(4x) ............................................................. (response #3 & #4)
CONTINUE......
D
Deleted member 4993
Guest
Why is substitution needed? Please refer to response #5 and #7.
skeeter
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Some excellent advice.
BigBeachBanana
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I'll call [imath]c=\cos(x),s=\sin(x)[/imath] for short.
[imath]c^4-6c^2s^2+s^4\underbrace{\red{+2c^2s^2-2c^2s^2}}_{\text{adding }0}\\[/imath]
The point is to have a perfect squares:
[math]\underbrace{c^4+s^4+2c^2s^2}_{\text{perfect squares}}-6c^2s^2-2c^2s^2= \underbrace{(c^2+s^2)^2-8c^2s^2}_{\text{difference of squares}}= (c^2+s^2-\sqrt{8}cs)(c^2+s^2+\sqrt{8}cs)=0[/math]Now set each factor equal to 0 and solve for the critical point. However, others have given you other approaches. Pick your poison.
