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Need some help in trigonometry!

Hello guys, I need some help in trigonometry! I am really not so good at it!😝

If \(x\cos(\theta)=y\cos(\theta+120^{\circ})=z\cos(\theta+240^{\circ})\)

Then find the value of \(xy+yz+zx\)

Note by Akshay Yadav
11 months, 1 week ago

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\(\begin{equation} \begin{split} x\cos (\theta) & = y \cos (\theta + 120^\circ) = z \cos (\theta + 240^\circ) \\ \Rightarrow \Re \left( x e^{i \theta} \right) & = \Re \left( y e^{i (\theta + 120^\circ)} \right) = \Re \left( z e^{i (\theta + 240^\circ)} \right) \\ \Re \left( x \right) & = \Re \left( y e^{i120^\circ} \right) = \Re \left( z e^{i 240^\circ} \right) \\ \Rightarrow x & = y \cos (120^\circ) = z \cos (240^\circ) \\ x & = -\frac{1}{2} y = -\frac{1}{2} z \\ \Rightarrow -2x & = y = z \end{split} \end{equation} \)

\(\Rightarrow xy +yz + zx = -2x^2 + 4x^2 - 2x^2 = \boxed{0}\) Chew-Seong Cheong · 11 months, 1 week ago

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@Chew-Seong Cheong Neat & fast :) Pulkit Gupta · 11 months, 1 week ago

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@Chew-Seong Cheong Thanks for telling me a new way to do this question. Akshay Yadav · 11 months, 1 week ago

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One possible approach is as follows.

From the given equations we have that \(y = \dfrac{\cos(\theta)}{\cos(\theta + 120)}x\) and that \(z = \dfrac{\cos(\theta)}{\cos(\theta + 240)}x.\)

Then \(xy + yz + xz = x^{2}*\left(\dfrac{\cos(\theta)}{\cos(\theta + 120)} + \dfrac{\cos^{2}(\theta)}{\cos(\theta + 120)\cos(\theta + 240)} + \dfrac{\cos(\theta)}{\cos(\theta + 240)}\right) =\)

\(x^{2} * \dfrac{\cos(\theta)}{\cos(\theta + 120)\cos(\theta + 240)}(\cos(\theta + 240) + \cos(\theta) + \cos(\theta + 120))\), (A).

But \(\cos(\theta + 120) = \cos(\theta)\cos(120) - \sin(\theta)\sin(120) = -\dfrac{1}{2}(\cos(\theta) + \sqrt{3}\sin(\theta))\)

and \(\cos(\theta + 240) = \cos(\theta)\cos(240) - \sin(\theta)\sin(240) = -\dfrac{1}{2}(\cos(\theta) - \sqrt{3}\sin(\theta)).\)

Thus \(\cos(\theta + 120) + \cos(\theta + 240) = -\cos(\theta)\), which makes expression (A) come out to \(\boxed{0}.\) Brian Charlesworth · 11 months, 1 week ago

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@Brian Charlesworth Thank you very much. Akshay Yadav · 11 months, 1 week ago

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