Math Problem Statement

(sin(𝜋/3)(tan𝜋/6) + csc(𝜋/4)^2

Solution

Let's solve the expression:

sin(π3)tan(π6)+csc2(π4)\sin\left(\frac{\pi}{3}\right) \cdot \tan\left(\frac{\pi}{6}\right) + \csc^2\left(\frac{\pi}{4}\right)

Step 1: Calculate each trigonometric function

  • sin(π3)=32\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}
  • tan(π6)=13\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}
  • csc(π4)=1sin(π4)=122=2\csc\left(\frac{\pi}{4}\right) = \frac{1}{\sin\left(\frac{\pi}{4}\right)} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}
  • csc2(π4)=(2)2=2\csc^2\left(\frac{\pi}{4}\right) = \left(\sqrt{2}\right)^2 = 2

Step 2: Substitute and simplify

The expression becomes: sin(π3)tan(π6)+csc2(π4)\sin\left(\frac{\pi}{3}\right) \cdot \tan\left(\frac{\pi}{6}\right) + \csc^2\left(\frac{\pi}{4}\right) Substitute the values: 3213+2\frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{3}} + 2

Simplifying the first term: 3213=12\frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{3}} = \frac{1}{2}

Now, add the second term: 12+2=12+42=52\frac{1}{2} + 2 = \frac{1}{2} + \frac{4}{2} = \frac{5}{2}

Final Answer:

52\boxed{\frac{5}{2}}

Would you like any additional details on this solution?

Here are 5 related questions you might find helpful:

  1. What is the general formula for trigonometric identities involving sine and tangent?
  2. How does the csc function relate to sine in trigonometry?
  3. How do you find the values of trigonometric functions at standard angles?
  4. What is the significance of π\pi-related angles in the unit circle?
  5. How do you simplify trigonometric expressions involving multiple functions?

Tip: Memorizing the exact values of trigonometric functions for common angles like π6\frac{\pi}{6}, π4\frac{\pi}{4}, and π3\frac{\pi}{3} can help speed up calculations!

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Trigonometric Functions
Simplification

Formulas

sin(𝜋/3) = √3/2
tan(𝜋/6) = 1/√3
csc(𝜋/4) = 1/sin(𝜋/4)
csc^2(𝜋/4) = 2

Theorems

Trigonometric Identities
Pythagorean Identities

Suitable Grade Level

Grades 10-12