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Exercise 3.4 · Q16

Q.If xcos⁡θ=ycos⁡(θ+2π3)=zcos⁡(θ+4π3)x\cos\theta = y\cos\left(\theta + \dfrac{2\pi}{3}\right) = z\cos\left(\theta + \dfrac{4\pi}{3}\right), find the value of xy+yz+zxxy+yz+zx.

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Step 1. Name the common value. Let xcos⁡θ=ycos⁡(θ+2π3)=zcos⁡(θ+4π3)=kx\cos\theta=y\cos(\theta+\tfrac{2\pi}3)=z\cos(\theta+\tfrac{4\pi}3)=k, so x=ksec⁡θx=k\sec\theta, y=ksec⁡(θ+2π3)y=k\sec(\theta+\tfrac{2\pi}3), z=ksec⁡(θ+4π3)z=k\sec(\theta+\tfrac{4\pi}3).

Step 2. Write xy+yz+zxxy+yz+zx over a common denominator.

xy+yz+zx=k2[cos⁡(θ+4π3)+cos⁡θ+cos⁡(θ+2π3)]cos⁡θcos⁡(θ+2π3)cos⁡(θ+4π3).xy+yz+zx=\frac{k^2\big[\cos(\theta+\tfrac{4\pi}3)+\cos\theta+\cos(\theta+\tfrac{2\pi}3)\big]}{\cos\theta\cos(\theta+\tfrac{2\pi}3)\cos(\theta+\tfrac{4\pi}3)}.

Step 3. Show the bracketed numerator is identically 00. By Identity 3.1, cos⁡(θ+2π3)=−12cos⁡θ−32sin⁡θ\cos(\theta+\tfrac{2\pi}3)=-\tfrac12\cos\theta-\tfrac{\sqrt3}2\sin\theta and cos⁡(θ+4π3)=−12cos⁡θ+32sin⁡θ\cos(\theta+\tfrac{4\pi}3)=-\tfrac12\cos\theta+\tfrac{\sqrt3}2\sin\theta (using cos⁡2π3=−12,sin⁡2π3=32\cos\tfrac{2\pi}3=-\tfrac12,\sin\tfrac{2\pi}3=\tfrac{\sqrt3}2 and cos⁡4π3=−12,sin⁡4π3=−32\cos\tfrac{4\pi}3=-\tfrac12,\sin\tfrac{4\pi}3=-\tfrac{\sqrt3}2). Adding all three: …

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