Find the Values of the Six Trigonometric Functions of Theta Based on Given Constraints

To find the values of the six trigonometric functions of theta (θ) where θ is in the second quadrant and sin(θ) = 35, we first note that the value of sin(θ) seems incorrect because the sine function can only take values between -1 and 1. However, if we consider that the problem might be intended to express sin(θ) in terms of a hypothetical triangle where the opposite side is 35 and the hypotenuse is 1 (a unique case), we can proceed with the calculations accordingly. We can use the Pythagorean theorem to find the adjacent side.

Let’s denote:

  • Opposite side = 35
  • Hypotenuse = 1

We use the Pythagorean theorem to find the adjacent side, represented as adj:
adj^2 + 35^2 = 1^2

This calculation seems invalid since we cannot have a valid triangle with these dimensions. Usually, we want sin(θ) to be a valid value between -1 and 1. If we were to use a reasonable trigonometric value, say sin(θ) = 0.35 instead, we can proceed with that to find the other functions in quadrant II.

Assuming sin(θ) = 0.35:

  • sin(θ) = 0.35
  • Since θ is in the second quadrant, cos(θ) will be negative and can be found using:

cos(θ) = -√(1 – sin²(θ))

Calculating this gives:

  • sin²(θ) = (0.35)² = 0.1225
  • cos(θ) = -√(1 – 0.1225) = -√(0.8775) ≈ -0.936

Now, we’ll calculate the other trigonometric functions:

  • tan(θ) = sin(θ) / cos(θ) = 0.35 / -0.936 ≈ -0.374
  • csc(θ) = 1/sin(θ) = 1/0.35 ≈ 2.857
  • sec(θ) = 1/cos(θ) = 1/-0.936 ≈ -1.070
  • cot(θ) = 1/tan(θ) = 1/-0.374 ≈ -2.673

So, the values of the six trigonometric functions for sin(θ) = 0.35 in quadrant II are approximately:

  • sin(θ) ≈ 0.35
  • cos(θ) ≈ -0.936
  • tan(θ) ≈ -0.374
  • csc(θ) ≈ 2.857
  • sec(θ) ≈ -1.070
  • cot(θ) ≈ -2.673

This completes the breakdown of finding the values of the six trigonometric functions of θ based on the given constraints.

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