Sin theta /x = cos theta /y then sin theta - cos theta=?

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Sin theta /x = cos theta /y then sin theta - cos theta=?

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Answer 1 · 2018-04-19T17:27:04

If $\frac{sin θ}{x} = \frac{cos θ}{y}$ $frac{ \sin theta }{x} = frac{ cos theta]{ y}$ then $sin θ - cos θ = \pm \frac{x - y}{\sqrt{x^{2} + y^{2}}}$ $\sin theta - cos theta = \pm frac {x - y}{sqrt{x^2+y^2}}$

Explanation:

$\frac{sin θ}{x} = \frac{cos θ}{y}$ $frac{ \sin theta }{x} = frac{ cos theta]{ y}$

$\frac{sin θ}{cos θ} = \frac{x}{y}$ $frac{ \sin theta}{\cos theta } = frac{x}{y}$

$tan θ = \frac{x}{y}$ $\tan \theta = x/y$

That's like a right triangle with opposite $x$ $x$ and adjacent $y$ $y$ so

$cos θ = \frac{\pm y}{\sqrt{x^{2} + y^{2}}}$ $cos theta = frac{\pm y}{sqrt{x^2 + y^2}$

$sin θ = tan θ cos θ$ $sin theta = \tan \theta \cos theta$

$sin θ - cos θ$ $\sin theta - cos theta$

$= tan θ cos θ - cos θ$ $= tan theta \cos theta - cos theta$

$= cos θ (tan θ - 1)$ $= \cos theta ( \tan theta - 1)$

$= \frac{\pm y}{\sqrt{x^{2} + y^{2}}} (\frac{x}{y} - 1)$ $= frac{\pm y}{sqrt{x^2 + y^2}} (x/y -1)$

$sin θ - cos θ = \pm \frac{x - y}{\sqrt{x^{2} + y^{2}}}$ $\sin theta - cos theta = \pm frac {x - y}{sqrt{x^2+y^2}}$

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Sin theta /x = cos theta /y then sin theta - cos theta=?

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