How do you find the exact value $sin (x + y)$ $sin(x+y)$ if $cos x = \frac{8}{17}, sin y = \frac{12}{37}$ $cosx=8/17, siny=12/37$ ?

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How do you find the exact value $sin (x + y)$ $sin(x+y)$ if $cos x = \frac{8}{17}, sin y = \frac{12}{37}$ $cosx=8/17, siny=12/37$ ?

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Answer 1 · 2016-11-22T08:30:58

The exact value is $= \frac{621}{629}$ $=621/629$

Explanation:

We know that

$sin (x + y) = sin x cos y + sin y cos x$ $sin(x+y)=sinxcosy+sinycosx$

If $cos x = \frac{8}{17}$ $cosx=8/17$ and $sin y = \frac{12}{37}$ $siny=12/37$

We can use, ${cos}^{2} x + {sin}^{2} x = 1$ $cos^2x+sin^2x=1$

and ${cos}^{2} y + {sin}^{2} y = 1$ $cos^2y+sin^2y=1$

To calculate $sin x$ $sinx$ and $cos y$ $cosy$

${sin}^{2} x = 1 - {cos}^{2} x = 1 - {(\frac{8}{17})}^{2} = \frac{225}{17^{2}}$ $sin^2x=1-cos^2x=1-(8/17)^2=225/17^2$

$sin x = \frac{15}{17}$ $sinx=15/17$

${cos}^{2} y = 1 - {sin}^{2} y = 1 - {(\frac{12}{37})}^{2} = \frac{1225}{37^{2}}$ $cos^2y=1-sin^2y=1-(12/37)^2=1225/37^2$

$cos y = \frac{35}{37}$ $cosy=35/37$

so,
$sin (x + y) = \frac{15}{17} \cdot \frac{35}{37} + \frac{12}{37} \cdot \frac{8}{17} = \frac{621}{629}$ $sin(x+y)=15/17*35/37+12/37*8/17=621/629$

Answer 2 · 2016-11-22T09:01:43

$sin (x + y) = \frac{621}{629}$ $sin(x+y)=621/629$ or $- \frac{429}{629}$ $-429/629$ depending on the quadrant in which sine and cosine lie.

Explanation:

Before we commence further, it may be mentioned that as $cos x = \frac{8}{17}$ $cosx=8/17$ , $x$ $x$ is in $Q 1$ $Q1$ or $Q 4$ $Q4$ i.e. $sin x$ $sinx$ could be positive or negative and as $sin y = \frac{12}{37}$ $siny=12/37$ , $y$ $y$ is in $Q 1$ $Q1$ or $Q 2$ $Q2$ i.e. $cos y$ $cosy$ could be positive or negative.

Hence four combinations for $(x + y)$ $(x+y)$ are there and for $sin (x + y) = sin x cos y + cos x sin y$ $sin(x+y)=sinxcosy+cosxsiny$ , there are four possibilities.

Now as $cos x = \frac{8}{17}$ $cosx=8/17$ , $sin x = \sqrt{1 - {(\frac{8}{17})}^{2}} = \sqrt{1 - \frac{64}{289}} = \sqrt{\frac{225}{289}} = \pm \frac{15}{17}$ $sinx=sqrt(1-(8/17)^2)=sqrt(1-64/289)=sqrt(225/289)=+-15/17$ and

as $sin y = \frac{12}{37}$ $siny=12/37$ , $cos y = \sqrt{1 - {(\frac{12}{37})}^{2}} = \sqrt{1 - \frac{144}{1369}} = \sqrt{\frac{1225}{1369}} = \pm \frac{35}{37}$ $cosy=sqrt(1-(12/37)^2)=sqrt(1-144/1369)=sqrt(1225/1369)=+-35/37$

Hence,

(1) when $x$ $x$ and $y$ $y$ are in $Q 1$ $Q1$

$sin (x + y) = \frac{15}{17} \times \frac{35}{37} + \frac{8}{17} \times \frac{12}{37} = \frac{525 + 96}{629} = \frac{621}{629}$ $sin(x+y)=15/17xx35/37+8/17xx12/37=(525+96)/629=621/629$

(2) when $x$ $x$ is in $Q 1$ $Q1$ and $y$ $y$ is in $Q 2$ $Q2$

$sin (x + y) = \frac{15}{17} \times \frac{- 35}{37} + \frac{8}{17} \times \frac{12}{37} = \frac{- 525 + 96}{629} = - \frac{429}{629}$ $sin(x+y)=15/17xx(-35)/37+8/17xx12/37=(-525+96)/629=-429/629$

(3) when $x$ $x$ is in $Q 4$ $Q4$ and $y$ $y$ is in $Q 2$ $Q2$

$sin (x + y) = \frac{- 15}{17} \times \frac{- 35}{37} + \frac{8}{17} \times \frac{12}{37} = \frac{525 + 96}{629} = \frac{621}{629}$ $sin(x+y)=(-15)/17xx(-35)/37+8/17xx12/37=(525+96)/629=621/629$

(4) when $x$ $x$ is in $Q 4$ $Q4$ and $y$ $y$ is in $Q 1$ $Q1$

$sin (x + y) = \frac{- 15}{17} \times \frac{35}{37} + \frac{8}{17} \times \frac{12}{37} = \frac{- 525 + 96}{629} = - \frac{429}{629}$ $sin(x+y)=(-15)/17xx35/37+8/17xx12/37=(-525+96)/629=-429/629$

Hence, $sin (x + y) = \frac{621}{629}$ $sin(x+y)=621/629$ or $- \frac{429}{629}$ $-429/629$

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