Find the exact value of the following trigonometric function.

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Find the exact value of the following trigonometric function.

sin 2pi/3
tan 5pi/6
csc 7pi/4
cos 5pi/3
csc 3pi/4
cot pi/6
sec 5pi/4
tan pi/2
If sin theta = 4/5 and theta is acute, find the exact value of the other five trigonometric functions. No decimals!

If sin theta = -Squareroot 3/2 and 3 pi/2 lessthanorequalto theta lessthanorequalto, find cos theta.

The Correct Answer and Explanation is:

Let’s solve each of the trigonometric expressions and provide a 300-word explanation for the final question.

Exact Values of Trigonometric Functions:

  1. sin(2π/3):
  • 2π/3 is in Quadrant II
  • Reference angle = π/3
  • sin(π/3) = √3/2 → In Quadrant II, sine is positive

Answer: √3/2

  1. tan(5π/6):
  • 5π/6 is in Quadrant II
  • Reference angle = π/6
  • tan(π/6) = 1/√3 → In Quadrant II, tangent is negative

Answer: –1/√3 or –√3/3

  1. csc(7π/4):
  • 7π/4 is in Quadrant IV
  • Reference angle = π/4
  • sin(π/4) = √2/2 → csc = 1/sin → In Quadrant IV, sine is negative

Answer: –√2

  1. cos(5π/3):
  • 5π/3 is in Quadrant IV
  • Reference angle = π/3
  • cos(π/3) = 1/2 → In Quadrant IV, cosine is positive

Answer: 1/2

  1. csc(3π/4):
  • 3π/4 is in Quadrant II
  • Reference angle = π/4
  • sin(π/4) = √2/2 → csc = 1/sin

Answer: √2

  1. cot(π/6):
  • cot = 1/tan
  • tan(π/6) = 1/√3

Answer: √3

  1. sec(5π/4):
  • 5π/4 is in Quadrant III
  • Reference angle = π/4
  • cos(π/4) = √2/2 → In Quadrant III, cosine is negative

Answer: –√2

  1. tan(π/2):
  • tan(π/2) is undefined because cos(π/2) = 0, and tan = sin/cos

Answer: Undefined

Given: sin(θ) = 4/5, θ is acute.

Use Pythagorean identity: sin⁡2(θ)+cos⁡2(θ)=1⇒(4/5)2+cos⁡2(θ)=1⇒16/25+cos⁡2(θ)=1⇒cos⁡2(θ)=9/25⇒cos⁡(θ)=3/5\sin^2(θ) + \cos^2(θ) = 1 \Rightarrow (4/5)^2 + \cos^2(θ) = 1 \Rightarrow 16/25 + \cos^2(θ) = 1 \Rightarrow \cos^2(θ) = 9/25 \Rightarrow \cos(θ) = 3/5

θ is acute, so cosine is positive.

Now find all 6 functions:

  • sin(θ) = 4/5
  • cos(θ) = 3/5
  • tan(θ) = sin/cos = 4/3
  • csc(θ) = 1/sin = 5/4
  • sec(θ) = 1/cos = 5/3
  • cot(θ) = cos/sin = 3/4

Now:

If sin(θ) = –√3/2 and 3π/2 ≤ θ ≤ 2π, find cos(θ).

Explanation (300+ words):

We are told that sin(θ) = –√3/2 and that θ lies between 3π/2 and 2π, which places θ in the fourth quadrant (Quadrant IV). In this quadrant, the sine function is negative, which matches the given value. The cosine function, however, is positive in this quadrant.

Let’s recall the Pythagorean identity: sin⁡2(θ)+cos⁡2(θ)=1\sin^2(θ) + \cos^2(θ) = 1

We substitute the known value of sin(θ): (−3/2)2+cos⁡2(θ)=1⇒3/4+cos⁡2(θ)=1⇒cos⁡2(θ)=1−3/4=1/4⇒cos⁡(θ)=±1/2(-\sqrt{3}/2)^2 + \cos^2(θ) = 1 \Rightarrow 3/4 + \cos^2(θ) = 1 \Rightarrow \cos^2(θ) = 1 – 3/4 = 1/4 \Rightarrow \cos(θ) = ±1/2

Now we decide on the correct sign. Since θ is in Quadrant IV, and cosine is positive in this quadrant, we choose the positive root.

Final Answer: cos(θ) = 1/2

To summarize, understanding which quadrant the angle lies in helps determine the correct sign of the trigonometric function. Even if a value such as cos²(θ) = 1/4 leads to both ±1/2, we must choose the sign based on the quadrant. In this case, the negative sine confirms the angle is in the fourth quadrant, and thus cosine must be positive. This approach ensures we not only compute the correct m

agnitude but also apply corr

ect directional signs according to the unit circle, a fundamental tool in trigonometry.

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