sin theta - cos theta =√6-√2÷2 then value of 4(sin^3 theta-cos^3 theta) ^2 is equal to
Answers
Solution :
Here it is given that :
sin theta - cos theta = ( √6 - √2 )/2
Now , cubing this on both the LHS and RHS
=> { sin theta - cos theta } ³ = { ( √6 - √2 )/2 }³
=> sin³ theta - 3sin² theta cos theta + 3 sin theta cos² theta - cos³ theta = [ √6 - √2 ]³/8
=> sin³ theta - cos ³ theta - 3sin theta cos theta ( sin theta - cos theta ) = [ 6^(3/2) - 3√12 ( 4 ) - 2^(3/2) ]/8
Now let us find the value of sin theta × cos theta .
sin theta - cos theta = ( √6 - √2 )/2
Squaring this
sin² theta + cos² theta - 2sin theta cos theta =[ 8 - 2√12 ] / 2
=> 4 - √12
Substituting this
sin³ theta - cos ³ theta - 3( 4 - √12 )( √6 - √2 )/2 = [ 6^(3/2) - 3√12 ( 4 ) - 2^(3/2) ]/8
=> sin ³ theta - cos ³ theta = [ 6^(3/2) - 3√12 ( 4 ) - 2^(3/2) ]/8 + 3( 4 - √12 )( √6 - √2 )/2
Lets consider the RHS now .
[ 6^(3/2) - 3√12 ( 4 ) - 2^(3/2) ]/8 + 3( 4 - √12 )( √6 - √2 )/2
Taking LCM
=>{ [ 6^(3/2) - 3√12 ( 4 ) - 2^(3/2) ] + 12 ( 4 - √12 )( √6 - √2 ) } /8
=> { [ 6^(3/2) - 3√12 ( 4 ) - 2^(3/2) ] + 12 [ 4√6 - 4√2 - √72 + √24 ] }/8
=> { [ 6^(3/2) - 3√12 ( 4 ) - 2^(3/2) ] + 12 [ 4√6 - 4√2 - 6√2 + 2√6 ] }/8
=> { [ 6^(3/2) - 3√12 ( 4 ) - 2^(3/2) ] + 12 [ 6√6 - 10√2 ] }/8
=> [ 6^(3/2) - 2^(3/2) - 12√12 + 72√6 - 120√2 ]/8
=>{ [ ³√36 ] - [ ³√4 ] - 24√3 + 72√6 - 120√2 } / 8
=> [ ³√36 - ³√4 ]/8 - 3√3 + 9√6 - 15√2.
This is the required answer .
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