Question

Some body please help in trigonometry class 11th problem...
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Answer 1

sin³∅ - cos³∅ =

( sin∅ - cos∅ )( sin² ∅ + cos²∅ + sin∅cos∅ )

( sin∅ - cos∅ )( 1 + sin∅cos∅ ) .... equation (1)

[ Formula used :-

a³ - b³ = ( a - b )( a² + b² + ab )

sin²∅ + cos²∅ = 1

]

sin∅ - cos∅ = (√6 - √2) / 2

Square both sides

( sin∅ - cos∅ )² = [ (√6 - √2) / 2 ]²

1 - 2sin∅cos∅ = ( 6 + 2 - 2√12 ) / 4

1 - 2sin∅cos∅ = ( 8 - 2√12 ) / 4

1 - ( 8 - 2√12 ) / 4 = 2sin∅cos∅

( 4 - 8 + 2√12 ) / 8 = sin∅cos∅

( √12 - 2 ) / 4 = sin∅cos∅

Put it in equation (1)

[ ( √6 - √2 ) / 2 ][ 1 + ( √12 - 2) / 4 ]

= 1 / √2

= sin³∅ - cos³∅

Now asked value is

4( sin³∅ - cos³∅ )²

4 ( 1/√2 )²

4/2

2

( I have emitted basic calculations like multiplying)

Answer

So, answer is 2.

Thank you.

Answer 2

The answer is 2 dear ..

I hope it will help you