What is the solution? If sin x=8/17, sin y= 15/17, and 0<x< π/2, 0<y<π/2; then x+y= π/2. Prove it!

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What is the solution? If sin x=8/17, sin y= 15/17, and 0<x< π/2, 0<y<π/2; then x+y= π/2. Prove it!

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Answer 1 · 2015-11-01T18:58:39

If $sin x = \frac{8}{17}$ $sinx=8/17$ then $cos x = \sqrt{1 - \frac{8^{2}}{17^{2}}} \Rightarrow cos x = \frac{15}{17}$ $cosx=sqrt(1-8^2/17^2)=>cosx=15/17$

and $sin y = \frac{15}{17}$ $siny=15/17$ then $cos y = \sqrt{1 - \frac{15^{2}}{17^{2}}} \Rightarrow cos y = \frac{8}{17}$ $cosy=sqrt(1-15^2/17^2)=>cosy=8/17$

So we have that

$sin (x + y) = sin x \cdot cos y + cos x \cdot sin y = \frac{8^{2} + 15^{2}}{17^{2}} = 1$ $sin(x+y)=sinx*cosy+cosx*siny=(8^2+15^2)/17^2=1$

Hence because $sin (x + y) = 1 \Rightarrow x + y = \frac{π}{2}$ $sin(x+y)=1=>x+y=pi/2$

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What is the solution? If sin x=8/17, sin y= 15/17, and 0<x< π/2, 0<y<π/2; then x+y= π/2. Prove it!

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