sinθ = O/H
cosθ = A/H
tanθ = O/A
cosθ*tanθ = A/H * O/A = AO/AH = O/H
Therefore, sinθ = cosθtanθ = O/H
(Notes: O=opposite, A=adjacent, H=hypotenuse)
because sin(2x) = 2sin(x)cos(x)
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
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°.
cosec(q)*cot(q)*cos(q) = 1/sin(q)*cot(q)*cos(q) = cot2(q)
If sin θ = tan θ, that means cos θ is 1 (since tan θ = (sin θ)/(cos θ)) (Usually in and equation a/b=a, b doesn't have to be 1 when a is 0, but cos θ = 1 if and only if sin θ = 0) The angles that satisfy cos θ = 1 is 2n(pi) (or 360n in degrees) When n is an integer. But if sin θ = tan θ = θ, the only answer is θ = 0. Because sin 0 is 0 and cos 0 is 1 and tan 0 is 0 The only answer would be when θ = 0.
Remember that tan = sin/cos. So your expression is sin/cos times cos. That's sin(theta).
There is a hint to how to solve this in what is required to be shown: a and b are both squared.Ifa cos θ + b sin θ = 8a sin θ - b cos θ = 5then square both sides of each to get:a² cos² θ + 2ab cos θ sin θ + b² sin² θ = 64a² sin² θ - 2ab sin θ cos θ + b² cos² θ = 25Now add the two together:a² cos² θ + a² sin² θ + b² sin² θ + b² cos² θ = 89→ a²(cos² θ + sin² θ) + b² (sin² θ + cos² θ) = 89using cos² θ + sin² θ = 1→ a² + b² = 89
'csc' = 1/sin'tan' = sin/cosSo it must follow that(cos) (csc) / (tan) = (cos) (1/sin)/(sin/cos) = (cos) (1/sin) (cos/sin) = (cos/sin)2
(Sin theta + cos theta)^n= sin n theta + cos n theta
The identity for tan(theta) is sin(theta)/cos(theta).
It's 1/2 of sin(2 theta) .
The expression "cot theta = 1.5 sin theta" can be rewritten using the definitions of trigonometric functions. Since cotangent is the reciprocal of tangent, we have cot(theta) = cos(theta) / sin(theta). Therefore, the equation becomes cos(theta) / sin(theta) = 1.5 sin(theta), leading to cos(theta) = 1.5 sin^2(theta). This relationship can be used to find specific values of theta that satisfy the equation.
The fourth Across the quadrants sin theta and cos theta vary: sin theta: + + - - cos theta: + - - + So for sin theta < 0, it's the third or fourth quadrant And for cos theta > 0 , it's the first or fourth quadrant. So for sin theta < 0 and cos theta > 0 it's the fourth quadrant
because sin(2x) = 2sin(x)cos(x)
The equation cannot be proved because of the scattered parts.
Remember use the Pythagorean Trig/ Identity. Sin^(2)(Theta) + Cos^(2)(Theta) = 1 Algebraically rearrange Sin^(2)(Theta) = 1 - Cos^(2)(Theta) Substitute Sin^(2)(Theta) = 1 - 0.65^(2) Factor Sin^(2)(Theta) = ( 1- 0.65 )( 1 + 0.65) Sin^(2)(Theta) = (0.35)(1.65) Sin^(2)(Theta) = 0.5775 Sin(Theta) = sqrt(0.5775) Sin(Theta) = 0.759934207.... Theta = Sun^(-1)(0.759934207...) Theta = 49.45839813 degrees.
The derivative of (sin (theta))^.5 is (cos(theta))/(2sin(theta))