Sin330.cos120+cos120.sin300
Answers
Answered by
1
The answer is given below :
Now,
sin330 cos120 + cos120 sin300
= sin(360 - 30) cos(180 - 60)
+ cos(180 - 60) sin(360 - 60)
= (- sin30)(- cos60) + (- cos60)(- sin60)
= sin30 cos60 + cos60 sin60
= (1/2 × 1/2) + (1/2 × √3/2)
= 1/4 + √3/4
= (1 + √3)/4
RULES :
1. sin(360 - θ) = - sinθ
2. cos(180 - θ) = - cosθ
3. sin30 = 1/2
4. cos60 = 1/2
5. sin60 = √3/2
Thank you for your question.
Now,
sin330 cos120 + cos120 sin300
= sin(360 - 30) cos(180 - 60)
+ cos(180 - 60) sin(360 - 60)
= (- sin30)(- cos60) + (- cos60)(- sin60)
= sin30 cos60 + cos60 sin60
= (1/2 × 1/2) + (1/2 × √3/2)
= 1/4 + √3/4
= (1 + √3)/4
RULES :
1. sin(360 - θ) = - sinθ
2. cos(180 - θ) = - cosθ
3. sin30 = 1/2
4. cos60 = 1/2
5. sin60 = √3/2
Thank you for your question.
Answered by
3
Hey......!!! here is ur answer......
_________________________
_________________________
We know that ,
sin(360-A)= -sinA
and cos(180-A) = -cosA
------------------------_-----------------------
Then according to the question,
=>sin330.cos120+cos120.sin330
=>sin(360-30).cos(180-60)+cos(180-60).sin(360-30)
=>-sin30.(-cos60)+(-cos60).(-sin30)
=>sin30.cos60+cos60.sin30
=>1/2×1/2+1/2×1/2
=>1/4+1/4
=>2/4=1/2
I hope it helps u.......☺️☺️☺️
THANK YOU..........✌️✌️✌️✌️
_________________________
_________________________
We know that ,
sin(360-A)= -sinA
and cos(180-A) = -cosA
------------------------_-----------------------
Then according to the question,
=>sin330.cos120+cos120.sin330
=>sin(360-30).cos(180-60)+cos(180-60).sin(360-30)
=>-sin30.(-cos60)+(-cos60).(-sin30)
=>sin30.cos60+cos60.sin30
=>1/2×1/2+1/2×1/2
=>1/4+1/4
=>2/4=1/2
I hope it helps u.......☺️☺️☺️
THANK YOU..........✌️✌️✌️✌️
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