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Ans to qsn no 15,
given , cos120°sin390° + cos330° cos150°
there for,
cos(90°+30°) sin(360° + 30°) + cos(360° -30°)cos(360° - 30°)cos(180° - 30°)
= - sin30°.sin30° + cos30°(-cos30°)
= - sin²30° - cos²30°
= - (sin²30° + cos²30°) ( because we knw that sin²A + cos²A = 1)
= - 1
plz give me brain mark dear.
given , cos120°sin390° + cos330° cos150°
there for,
cos(90°+30°) sin(360° + 30°) + cos(360° -30°)cos(360° - 30°)cos(180° - 30°)
= - sin30°.sin30° + cos30°(-cos30°)
= - sin²30° - cos²30°
= - (sin²30° + cos²30°) ( because we knw that sin²A + cos²A = 1)
= - 1
plz give me brain mark dear.
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