Question

plzzz solve...
$( \cos(0 ) + \sin(45) + \sin(30) )( \sin(90) - \cos(45) + \cos(60) )$

Answer 1

Hi ,

We know that ,

Cos0° = 1 ,

Sin45° = 1/√2 ,

Sin30° = 1/2 ,

Sin 90° = 1 ,

Cos 45° = 1/√2 ,

Cos 60° = 1/2 ;

Now ,

(Cos0 + sin45+sin30)(sin90-cos45+cos60)

= ( 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

= ( 3/2 )² - ( 1/√2 )²

= 9/4 - 1/2

= ( 9 - 2 )/4

= 7/4

I hope this helps you.

: )