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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

1/4
3/4
1/8
3/8

The correct answer is: 1/8

We have $fraction numerator 7 pi over denominator 8 end fraction equals cos invisible function application open parentheses pi minus fraction numerator pi over denominator 8 end fraction close parentheses equals negative cos invisible function application fraction numerator pi over denominator 8 end fraction$
And $cos invisible function application fraction numerator 5 pi over denominator 8 end fraction equals cos invisible function application open parentheses pi minus fraction numerator 3 pi over denominator 8 end fraction close parentheses equals negative cos invisible function application fraction numerator 3 pi over denominator 8 end fraction$
$therefore L. H. S. equals open parentheses 1 plus cos invisible function application fraction numerator pi over denominator 8 end fraction close parentheses open parentheses 1 plus cos invisible function application fraction numerator 3 pi over denominator 8 end fraction close parentheses open parentheses 1 minus cos invisible function application fraction numerator 3 pi over denominator 8 end fraction close parentheses open parentheses 1 minus cos invisible function application fraction numerator pi over denominator 8 end fraction close parentheses$
$equals open parentheses 1 minus cos to the power of 2 end exponent invisible function application fraction numerator pi over denominator 8 end fraction close parentheses open parentheses 1 minus cos to the power of 2 end exponent invisible function application fraction numerator 3 pi over denominator 8 end fraction close parentheses$
$equals sin to the power of 2 end exponent invisible function application fraction numerator pi over denominator 8 end fraction sin to the power of 2 end exponent invisible function application fraction numerator 3 pi over denominator 8 end fraction$
$equals fraction numerator 1 over denominator 4 end fraction open parentheses 2 sin to the power of 2 end exponent invisible function application fraction numerator pi over denominator 8 end fraction close parentheses open parentheses 2 sin to the power of 2 end exponent invisible function application fraction numerator 3 pi over denominator 8 end fraction close parentheses$
$equals fraction numerator 1 over denominator 4 end fraction open square brackets open parentheses 1 minus cos invisible function application fraction numerator pi over denominator 4 end fraction close parentheses open parentheses 1 minus cos invisible function application fraction numerator 3 pi over denominator 4 end fraction close parentheses close square brackets$ $open square brackets because 1 minus cos invisible function application theta equals 2 sin to the power of 2 end exponent invisible function application fraction numerator theta over denominator 2 end fraction close square brackets$
$equals fraction numerator 1 over denominator 4 end fraction open square brackets open parentheses 1 minus fraction numerator 1 over denominator square root of 2 end fraction close parentheses open parentheses 1 plus fraction numerator 1 over denominator square root of 2 end fraction close parentheses close square brackets equals fraction numerator 1 over denominator 4 end fraction open parentheses 1 minus fraction numerator 1 over denominator 2 end fraction close parentheses equals fraction numerator 1 over denominator 8 end fraction equals fraction numerator 1 over denominator 4 end fraction open parentheses 1 minus fraction numerator 1 over denominator 2 end fraction close parentheses equals fraction numerator 1 over denominator 8 end fraction equals R. H. S.$

Related Questions to study

triangle $A B C comma fraction numerator sin invisible function application A plus sin invisible function application B plus sin invisible function application C over denominator sin invisible function application A plus sin invisible function application B minus sin invisible function application C end fraction$ is equal to

triangle $A B C comma fraction numerator sin invisible function application A plus sin invisible function application B plus sin invisible function application C over denominator sin invisible function application A plus sin invisible function application B minus sin invisible function application C end fraction$ is equal to

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parallel

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$alpha$ is a root of $25 cos to the power of 2 end exponent invisible function application theta plus 5 cos invisible function application theta minus 12 equals 0 comma fraction numerator pi over denominator 2 end fraction less than alpha less than pi comma$ then $blank sin invisible function application 2 alpha blank$ is equal to

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$x y to the power of 2 end exponent equals 4 blank a n d log subscript 3 end subscript invisible function application open parentheses log subscript 2 end subscript invisible function application x close parentheses plus log subscript 1 divided by 3 end subscript invisible function application open parentheses log subscript 1 divided by 2 end subscript invisible function application y close parentheses equals 1 comma$ then $x$ equals

$I$ is the incentre of a triangle $A B C comma$ then the ratio $I A colon I B colon I C$ is equal to

$I$ is the incentre of a triangle $A B C comma$ then the ratio $I A colon I B colon I C$ is equal to

parallel

$log subscript 3 end subscript invisible function application left curly bracket 5 plus 4 log subscript 3 end subscript invisible function application open parentheses x minus 1 close parentheses right curly bracket blank equals 2 comma$ then $x$ is equal to

$log subscript 3 end subscript invisible function application left curly bracket 5 plus 4 log subscript 3 end subscript invisible function application open parentheses x minus 1 close parentheses right curly bracket blank equals 2 comma$ then $x$ is equal to

$log subscript y end subscript invisible function application x plus log subscript x end subscript invisible function application y equals 1 comma blank x to the power of 2 end exponent plus y equals 12 comma blank$ then the value of $x y$ is

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parallel

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parallel

$log subscript 2 end subscript invisible function application x plus log subscript 2 end subscript invisible function application y greater or equal than 6 comma$ then the least value of $x plus y$ is

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t $a greater than 1$ be a real number. Then the number of roots equation $a to the power of 2 log subscript 2 end subscript invisible function application x end exponent equals 5 plus 4 x to the power of log subscript 2 end subscript invisible function application a end exponent$ has

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parallel

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