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Let $alpha blank a n d blank beta$ be such that $pi less than alpha minus beta less than 3 pi.$ If $sin invisible function application alpha plus sin invisible function application beta equals negative fraction numerator 21 over denominator 65 end fraction blank a n d cos invisible function application alpha plus cos invisible function application beta equals negative fraction numerator 17 over denominator 65 end fraction comma$ then the value of $cos invisible function application fraction numerator alpha minus beta over denominator 2 end fraction$ is

$- \frac{3}{\sqrt{130}}$
$\frac{3}{\sqrt{130}}$
$\frac{6}{65}$
$- \frac{6}{65}$

The correct answer is: $negative fraction numerator 3 over denominator square root of 130 end fraction$

$W e blank h a v e sin invisible function application alpha plus sin invisible function application beta equals negative fraction numerator 21 over denominator 65 end fraction blank$
$cos invisible function application alpha plus cos invisible function application beta equals negative fraction numerator 17 over denominator 65 end fraction$
Squaring Eq. (i), we get $sin to the power of 2 end exponent invisible function application alpha plus sin to the power of 2 end exponent invisible function application beta plus 2 sin invisible function application alpha sin invisible function application beta equals open parentheses fraction numerator 21 over denominator 65 end fraction close parentheses to the power of 2 end exponent$
Squaring Eq. (ii), we get $cos to the power of 2 end exponent invisible function application alpha plus cos to the power of 2 end exponent invisible function application beta plus 2 cos invisible function application alpha cos invisible function application beta equals open parentheses fraction numerator 27 over denominator 65 end fraction close parentheses to the power of 2 end exponent$
Adding Eqs. (iii) and (iv), we get $2 plus 2 cos invisible function application open parentheses alpha minus beta close parentheses equals fraction numerator 1 over denominator open parentheses 65 close parentheses to the power of 2 end exponent end fraction open square brackets open parentheses 27 close parentheses to the power of 2 end exponent plus open parentheses 21 close parentheses to the power of 2 end exponent close square brackets equals fraction numerator 1 over denominator open parentheses 65 close parentheses to the power of 2 end exponent end fraction open parentheses 729 plus 441 close parentheses$
$rightwards double arrow 2 plus 2 blank cos invisible function application open parentheses alpha minus beta close parentheses equals fraction numerator 1 over denominator open parentheses 65 close parentheses to the power of 2 end exponent end fraction open parentheses 1170 close parentheses equals fraction numerator 18 over denominator 65 end fraction$
$rightwards double arrow 1 plus cos invisible function application open parentheses alpha minus beta close parentheses equals fraction numerator 9 over denominator 65 end fraction$
$rightwards double arrow 2 cos to the power of 2 end exponent invisible function application fraction numerator alpha minus beta over denominator 2 end fraction equals fraction numerator 9 over denominator 65 end fraction$
$rightwards double arrow cos invisible function application fraction numerator alpha minus beta over denominator 2 end fraction equals negative fraction numerator 3 over denominator square root of 130 end fraction blank open square brackets because pi less than alpha minus beta less than 3 pi rightwards double arrow fraction numerator pi over denominator 2 end fraction less than fraction numerator alpha minus beta over denominator 2 end fraction less than fraction numerator 3 pi over denominator 2 end fraction rightwards double arrow cos invisible function application open parentheses fraction numerator alpha minus beta over denominator 2 end fraction close parentheses less than 0 close square brackets$

Related Questions to study

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One of the general solutions of $4 sin to the power of 4 end exponent invisible function application x plus cos to the power of 4 end exponent invisible function application x equals 1$ is

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If $sin invisible function application theta subscript 1 end subscript plus sin invisible function application theta subscript 2 end subscript plus sin invisible function application theta subscript 3 end subscript equals 3 comma$ then $cos invisible function application theta subscript 1 end subscript plus cos invisible function application theta subscript 2 end subscript plus cos invisible function application theta subscript 3 end subscript$ is equal to

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parallel

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