Answer:
To find: Value of cos 2220°
We have,
cos (-2220 ) = cos 2220°
[∵ cos(-θ) = cos θ]
= cos [2160 + 60°]
= cos [360° × 6 + 60°]
= cos 60°
[Clearly, 2220° is in I Quadrant and the multiple of 360° is even]
Answer:
To find: Value of cos 2220°
We have,
cos (-2220 ) = cos 2220°
[∵ cos(-θ) = cos θ]
= cos [2160 + 60°]
= cos [360° × 6 + 60°]
= cos 60°
[Clearly, 2220° is in I Quadrant and the multiple of 360° is even]