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What trigonometric value is equal to cos 62?
The solution is found by applying the definition of complementary trig functions: Cos (&Theta) = sin (90°-&Theta) cos (62°) = sin (90°-62°) Therefore the solution is sin 28°.
What is the exact value of cos theta if csc theta -4 with theta in quadrant III?
csc θ = 1/sin θ → sin θ = -1/4 cos² θ + sin² θ = 1 → cos θ = ± √(1 - sin² θ) = ± √(1 - ¼²) = ± √(1- 1/16) = ± √(15/16) = ± (√15)/4 In Quadrant III both cos and sin are negative → cos θ= -(√15)/4
How do you simplify csc theta cot theta cos theta?
cosec(q)*cot(q)*cos(q) = 1/sin(q)*cot(q)*cos(q) = cot2(q)
If sin(36 plus theta) cos(16 plus theta) find theta?
cos(α) = sin(90° - α) → cos(16° + θ) = sin(90° - (16° + θ)) = sin(74° - θ) → sin(36° + θ) = cos(16° + θ) → sin((36° + θ) = sin(74° - θ) → 36° + θ = 74° - θ → 2θ = 38° → θ = 19° → θ = 19 °+ 180°n for n= 0, 1, 2, ...
What is cos theta minus cos theta times sin squared theta?
cos(t) - cos(t)*sin2(t) = cos(t)*[1 - sin2(t)] But [1 - sin2(t)] = cos2(t) So, the expression = cos(t)*cos2(t) = cos3(t)
