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 $xy+yz+zx$

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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

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

Chew-Seong Cheong - 5 years, 5 months ago

Neat & fast :)

Pulkit Gupta - 5 years, 5 months ago

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

Akshay Yadav - 5 years, 5 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 $\boxed{0}.$

Brian Charlesworth - 5 years, 5 months ago

Thank you very much.

Akshay Yadav - 5 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$