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$fraction numerator sin invisible function application x over denominator sin invisible function application y end fraction equals fraction numerator 1 over denominator 2 end fraction comma fraction numerator cos invisible function application x over denominator cos invisible function application y end fraction equals fraction numerator 3 over denominator 2 end fraction blank w h e r e blank x comma blank y blank element of open parentheses 0 comma fraction numerator pi over denominator 2 end fraction close parentheses comma blank t h e n blank t h e blank v a l u e blank o f blank left parenthesis x plus y right parenthesis blank i s blank e q u a l blank t o$

$\sqrt{13}$
$\sqrt{14}$
$\sqrt{17}$
$\sqrt{15}$

The correct answer is: $square root of 15$

$fraction numerator sin invisible function application x over denominator sin invisible function application y end fraction equals fraction numerator 1 over denominator 2 end fraction comma fraction numerator cos invisible function application x over denominator cos invisible function application y end fraction equals fraction numerator 3 over denominator 2 end fraction rightwards double arrow fraction numerator tan invisible function application x over denominator tan invisible function application y end fraction equals fraction numerator 1 over denominator 3 end fraction rightwards double arrow tan invisible function application open parentheses x plus y close parentheses equals fraction numerator tan invisible function application x plus tan invisible function application y over denominator 1 minus tan invisible function application x tan invisible function application y end fraction equals fraction numerator 4 tan invisible function application x over denominator 1 minus 3 tan to the power of 2 end exponent invisible function application x end fraction$
$A l s o sin invisible function application y equals 2 sin invisible function application x comma cos invisible function application y equals fraction numerator 2 over denominator 3 end fraction cos invisible function application x$
$rightwards double arrow sin to the power of 2 end exponent invisible function application y plus cos to the power of 2 end exponent invisible function application y equals 4 sin to the power of 2 end exponent invisible function application x plus fraction numerator 4 cos to the power of 2 end exponent invisible function application x over denominator 9 end fraction equals 1$
$rightwards double arrow 36 tan to the power of 2 end exponent invisible function application x plus 4 equals 9 sec to the power of 2 end exponent invisible function application x equals 9 open parentheses 1 plus tan to the power of 2 end exponent invisible function application x close parentheses$
$rightwards double arrow 27 tan to the power of 2 end exponent invisible function application x equals 5$
$rightwards double arrow tan invisible function application x equals fraction numerator square root of 5 over denominator 3 square root of 3 end fraction$
$rightwards double arrow tan invisible function application open parentheses x plus y close parentheses equals fraction numerator fraction numerator 4 square root of 5 over denominator 3 square root of 3 end fraction over denominator 1 minus fraction numerator 15 over denominator 27 end fraction end fraction equals fraction numerator 4 square root of 5 cross times 27 over denominator 12 cross times 3 square root of 3 end fraction equals square root of 15$

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r $blank n element of Z comma blank$ the general solution of $open parentheses square root of 3 minus 1 close parentheses sin invisible function application theta plus open parentheses square root of 3 plus 1 close parentheses cos invisible function application theta equals 2 blank$ is $open parentheses n element of Z close parentheses$

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e value of $open parentheses 1 plus cos invisible function application fraction numerator pi over denominator 8 end fraction close parentheses open parentheses 1 plus cos invisible function application fraction numerator 3 pi over denominator 8 end fraction close parentheses open parentheses 1 plus cos invisible function application fraction numerator 5 pi over denominator 8 end fraction close parentheses open parentheses 1 plus cos invisible function application fraction numerator 7 pi over denominator 8 end fraction close parentheses$ is

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