# hard!

• Jun 12th 2007, 01:22 AM
calculus_jy
hard!
can u help me solve this question

If xcos θ + y sinθ = a cos 2θ and;
xsinθ - ycos θ = 2a sin 2θ
prove that (x+y)2/3 + (x-y)2/3 = 2a2/3
• Jun 12th 2007, 04:56 AM
Soroban
Hello, calculus_jy!

This is The Problem from Hell . . . (coming to a theater near you)

Quote:

If . $\begin{array}{cc}x\cos\theta + y\sin\theta\:=\:a\cos2\theta & [1]\\ x\sin\theta - y\cos\theta\:=\:2a\sin2\theta & [2]\end{array}$

prove that: . $(x+y)^{\frac{2}{3}} + (x-y)^{\frac{2}{3}} \:=\:2a^{\frac{2}{3}}$

Multiply [1] by $\cos\theta:\;\;x\cos^2\!\theta + y\sin\theta\cos\theta \:=\:a\cos2\theta\cos\theta$
Multiply [2] by $\sin\theta:\;\;x\sin^2\!\theta - y\sin\theta\cos\theta\:=\:2a\sin2\theta\sin\theta$

Add: . $x\cos^2\!\theta + x\sin^2\!\theta \:=\:a\cos2\theta\cos\theta + 2a\sin2\theta\cos\theta$

. . . . $x(\cos^2\!\theta + \sin^2\!\theta) \:=\:a(\cos^2\!\theta - \sin^2\!\theta)\cos\theta + 2a(2\sin\theta\cos\theta)\sin\theta$

. . . . . . . . . . . . . $x \:=\:a\cos^3\!\theta - a\sin^2\!\theta\cos\theta + 4a\sin^2\!\theta\cos\theta$

. . . . . . . . . . . . . $x \:=\:a\cos^3\!\theta + 3a\sin^2\!\theta\cos\theta$

In a similar fashion solve for $y:\;\;y \:=\:-3a\cos^2\!\theta\sin\theta - a\sin^3\!\theta$

We have: . $x + y \;=\;(a\cos^3\!\theta + 3a\cos\theta\sin^2\!\theta) + (-3a\cos^2\!\theta\sin\theta - a\sin^3\!\theta)$
. . . . . . . . . . . . $= \;a(\cos^3\!\theta - 3\cos^2\!\theta\sin\theta + 3\cos\theta\sin^2\!\theta - \sin^3\!\theta)$
. . . . . . . . . . . . $= \;a(\cos\theta - \sin\theta)^3$

We have: . $x - y \;=\;(a\cos^3\!\theta + 3a\cos\theta\sin^2\!\theta) - (-3a\cos^2\!\theta\sin\theta - a\sin^3\!\theta)$
. . . . . . . . . . . . $= \;a(\cos^3\!\theta + 3\cos^2\!\theta\sin\theta + 3\cos\theta\sin^2\!\theta + \sin^3\!\theta)$
. . . . . . . . . . . . $= \;a(\cos\theta + \sin\theta)^3$

Then: . $\begin{array}{cccc}x + y \:=\:a(\cos\theta - \sin\theta)^3 & \Rightarrow & (x + y)^{\frac{2}{3}} \:=\:a^{\frac{2}{3}}(\cos\theta - \sin\theta)^2 & [3]\\
x - y \:=\:a(\cos\theta + \sin\theta)^3 & \Rightarrow & (x - y)^{\frac{2}{3}} \:=\:a^{\frac{2}{3}}(\cos\theta + \sin\theta)^2 & [4]\end{array}$

. . $(x+y)^{\frac{2}{3}} + (x - y)^{\frac{2}{3}} \;=\;a^{\frac{2}{3}}\bigg[(\cos\theta - \sin\theta)^2 + (\cos\theta + \sin\theta)^2\bigg]$

. . . . . . . . . . . . . . $= \;a^{\frac{2}{3}}\left[\cos^2\!\theta - 2\sin\theta\cos\theta + \sin^2\!\theta + \cos^2\!\theta + 2\sin\theta\cos\theta + \sin^2\!\theta\right]$

. . . . . . . . . . . . . . $= \;a^{\frac{2}{3}}\left[2\cos^2\!\theta + 2\sin^2\!\theta\right]$

. . . . . . . . . . . . . . $= \;2a^{\frac{2}{3}}\left[\cos^2\!\theta + \sin^2\!\theta\right]$

Therefore: . $(x + y)^{\frac{2}{3}} + (x - y)^{\frac{2}{3}} \;=\;2a^{\frac{2}{3}}$ . . . . ta-DAA!!

• Jun 12th 2007, 10:45 PM
calculus_jy
Great thx
i really appreciate ur help, and u said it is a problem from hell, just wondering how u manage to start solving problems lyk this, besides practicing and using ur brain, can u offer me few tips??

Also i hav 2 other problems from hell, i will post it in the geometry section and wonder if u can help me with them, they took me ages!