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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
1 year, 10 months ago

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

$$\Rightarrow xy +yz + zx = -2x^2 + 4x^2 - 2x^2 = \boxed{0}$$

- 1 year, 10 months ago

Neat & fast :)

- 1 year, 10 months ago

Thanks for telling me a new way to do this question.

- 1 year, 10 months ago

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 $$\boxed{0}.$$

- 1 year, 10 months ago

Thank you very much.

- 1 year, 10 months ago