Question

sin45°×cos65°+sin135°×cos115°

Answer 1

Hey !!

sin45° × cos65° + sin135° ×cos115° .

=> sin45° ×cos(90°-45°) + sin(180°-45°) ×cos(90°+45° )

=> sin45° ×sin45° + cos45° × sin45° ×(-sin45° )

=> sin²45° - cos²45°

=> (1/√2)² - (1/√2)²

=> 1/2 - 1/2

=> 1 - 1/2 = 0/2 = 0

******************************

Hope it helps you !!!

@Rajukumar111

Answer 2

Answer:

The answer is 0.

Explanation :

sin 45° × cos 65° + sin 135° × cos 115°

= sin 45° × cos 65° + sin (180 - 45)° × cos (180 - 65)°

= sin 45° × cos 65° + ( sin 45°) × ( - cos 65° )

( We know that,  sin (180 - x) = sin x and cos (180-x) = - cos x )

= sin 45° × cos 65° -  sin 45° × cos 65°

= 0

Hence, sin 45° × cos 65° + sin 135° × cos 115° = 0