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You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.

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It is +/- 0.76

Q: If cos and theta 0.65 what is the value of sin and theta?

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(/) = theta sin 2(/) = 2sin(/)cos(/)

sin/cos

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

-0.5736

Suppose triangle ABC is right angled at C. Suppose you are given that the angle at B is theta. Thenif you know the length of AB (the hypotenuse), thenBC = AB*cos(theta) andAC = AB*sin(theta)if you know the length of BC, thenAB = BC/cos(theta) andAC = BC*tan(theta)if you know the length of AC, thenAB= AC/sin(theta) andBC = AC/tan(theta)

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Remember that tan = sin/cos. So your expression is sin/cos times cos. That's sin(theta).

'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).

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°.

It's 1/2 of sin(2 theta) .

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.

The derivative of (sin (theta))^.5 is (cos(theta))/(2sin(theta))

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

Let 'theta' = A [as 'A' is easier to type] sec A - 1/(sec A) = 1/(cos A) - cos A = (1 - cos^2 A)/(cos A) = (sin^2 A)/(cos A) = (tan A)*(sin A) Then you can swap back the 'A' with theta