Math Problem Statement

A)csc(3pi/4), b) tan(-5pi/6), c) sec(-7pi/3)

Solution

Let's solve each of the trigonometric functions:

a) csc(3π4)\csc\left(\frac{3\pi}{4}\right)

  • The cosecant function is the reciprocal of the sine:
    csc(θ)=1sin(θ)\csc(\theta) = \frac{1}{\sin(\theta)}
  • First, determine sin(3π4)\sin\left(\frac{3\pi}{4}\right). The angle 3π4\frac{3\pi}{4} lies in the second quadrant, where sine is positive.
  • sin(3π4)=sin(ππ4)=sin(π4)=22\sin\left(\frac{3\pi}{4}\right) = \sin\left(\pi - \frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}

Thus: csc(3π4)=122=22=2\csc\left(\frac{3\pi}{4}\right) = \frac{1}{\frac{\sqrt{2}}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}

b) tan(5π6)\tan\left(-\frac{5\pi}{6}\right)

  • The tangent function is periodic with a period of π\pi, and for negative angles, we reflect across the x-axis.
  • tan(5π6)=tan(5π6)\tan\left(-\frac{5\pi}{6}\right) = -\tan\left(\frac{5\pi}{6}\right)
  • The angle 5π6\frac{5\pi}{6} is in the second quadrant, where tangent is negative.
    tan(5π6)=tan(ππ6)=tan(π6)=13\tan\left(\frac{5\pi}{6}\right) = \tan\left(\pi - \frac{\pi}{6}\right) = -\tan\left(\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}}
  • Therefore: tan(5π6)=13\tan\left(-\frac{5\pi}{6}\right) = \frac{1}{\sqrt{3}}

c) sec(7π3)\sec\left(-\frac{7\pi}{3}\right)

  • The secant function is the reciprocal of the cosine: sec(θ)=1cos(θ)\sec(\theta) = \frac{1}{\cos(\theta)}
  • To simplify sec(7π3)\sec\left(-\frac{7\pi}{3}\right), use the periodicity of cosine (period 2π2\pi).
  • 7π3-\frac{7\pi}{3} can be rewritten as:
    7π3+2π=7π3+6π3=π3-\frac{7\pi}{3} + 2\pi = -\frac{7\pi}{3} + \frac{6\pi}{3} = -\frac{\pi}{3}
  • sec(π3)=sec(π3)\sec\left(-\frac{\pi}{3}\right) = \sec\left(\frac{\pi}{3}\right) (cosine is an even function).
  • cos(π3)=12\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}

Thus: sec(π3)=112=2\sec\left(-\frac{\pi}{3}\right) = \frac{1}{\frac{1}{2}} = 2

Final answers:

a) csc(3π4)=2\csc\left(\frac{3\pi}{4}\right) = \sqrt{2}
b) tan(5π6)=13\tan\left(-\frac{5\pi}{6}\right) = \frac{1}{\sqrt{3}}
c) sec(7π3)=2\sec\left(-\frac{7\pi}{3}\right) = 2


Would you like further details or have any questions on these?
Here are some related questions:

  1. What is the value of sin(5π6)\sin\left(\frac{5\pi}{6}\right)?
  2. How do you find the period of trigonometric functions like secant or cosecant?
  3. Why is tangent negative in the second quadrant?
  4. What is the relationship between cotangent and tangent?
  5. Can you explain how trigonometric identities apply to negative angles?

Tip: When simplifying trigonometric functions with large angles, remember to reduce them using the periodicity of the function.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Reciprocal Functions
Periodic Functions
Angle Reduction
Even and Odd Functions

Formulas

csc(θ) = 1/sin(θ)
tan(θ) = sin(θ)/cos(θ)
sec(θ) = 1/cos(θ)

Theorems

Reciprocal Trigonometric Functions
Periodicity of Trigonometric Functions
Reflection of Trigonometric Functions for Negative Angles

Suitable Grade Level

Grades 10-12