Question

Evaluate each of the following (cos 0° + sin 45° + sin 30°) (sin 90° − cos 45° + cos 60°)

Answer 1

SOLUTION :

Given :

(cos 0° + sin 45°+ sin 30°) (sin 90° - cos 45° + cos 60°)

= (1+1/√2 +1/2) (1- 1/√2 + 1/2)

[cos 0° = 1 , sin 45° = 1/√2 , sin 30°= ½, sin 90° = 1 , cos 0° = 1 , cos 45° = 1/√2 ,  cos 60°= ½]

= (1 + ½ + 1/√2)  (1 + ½ - 1/√2)

[ By rearranging the terms ]

= ((2+1)/2 +1/√2) ((2 + 1)/2 - 1/√2)  

[ Taking L.C.M of 1 & 2 is 2]

= (3/2 + 1/√2)  (3/2 - 1/√2)  

= (3√2 + 2)/2√2 (3√2 - 2)/2√2

[Taking L.C.M of 2 & √2 is 2√2]

= [ (3√2 + 2)  (3√2 - 2)] / 2√2 × 2√2

= (3√2)² - (2)² / 4× 2

[(a+b) (a-b) = a² - b²]

=( 9× 2 - 4)/8

= (18 - 4)/8

= 14/8  

= 14÷2 / 8÷2

= 7/4

(cos 0° + sin 45°+ sin 30°) (sin 90° - cos 45° + cos 60°) = 7/4

Hence, (cos 0° + sin 45°+ sin 30°) (sin 90° - cos 45° + cos 60°) = 7/4

HOPE THIS ANSWER WILL HELP YOU.

Answer 2

Answer :-


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(cos 0° + sin 45° + sin 30°) (sin 90° − cos 45° + cos 60°)

=> ( 1 + 1 / √2 + 1 / 2 ) ( 1 - 1 / √2 + 1 / 2 )

=> { ( 1 + 1 / 2 ) + ( 1 / √2 ) } { ( 1 + 1 / 2 ) - ( 1 / √2 ) }

=> ( 1 + 1 / 2 )² - ( 1 / √2 )²

[ • a² - b² = ( a + b ) ( a - b ) , where , a = ( 1 + 1 / 2 ) and b = ( 1 / √2 ) ]

=> { ( 1 )² + 2 . 1 . 1 / 2 + ( 1 / 2 )² } - ( 1 / 2 )

=> ( 1 + 1 + 1 / 4 ) - 1 / 2

=> 2 + 1 / 4 - 1 / 2

=> ( 8 + 1 - 2 ) / 4

=> 7 / 4


•°• The required Answer is 7 / 4


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Since ,

cos0° = 1

cos45° = 1 / √2

cos60° = 1 /2

sin90° = 1

sin45° = 1 / √2

sin30° = 1 / 2

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