Math Problem Statement

If ๐œƒ = ๐œ‹/3, find the value of each expression. (Round your answers to three decimal places, if necessary.) sin2(๐œƒ) =

sin(๐œƒ^2) =

Solution

Given that ฮธ=ฯ€3\theta = \frac{\pi}{3}, we need to find the values of the following expressions:

  1. sinโก2(ฮธ)\sin^2(\theta)
  2. sinโก(ฮธ2)\sin(\theta^2)

Step 1: Calculating sinโก2(ฮธ)\sin^2(\theta)

First, calculate sinโก(ฮธ)\sin(\theta) for ฮธ=ฯ€3\theta = \frac{\pi}{3}.

We know that:

sinโก(ฯ€3)=32โ‰ˆ0.866\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \approx 0.866

Now, square this value to find sinโก2(ฮธ)\sin^2(\theta):

sinโก2(ฯ€3)=(0.866)2=0.750\sin^2\left(\frac{\pi}{3}\right) = \left( 0.866 \right)^2 = 0.750

Step 2: Calculating sinโก(ฮธ2)\sin(\theta^2)

Next, calculate ฮธ2\theta^2, which is:

ฮธ2=(ฯ€3)2=ฯ€29โ‰ˆ1.096\theta^2 = \left(\frac{\pi}{3}\right)^2 = \frac{\pi^2}{9} \approx 1.096

Now, compute sinโก(ฮธ2)\sin(\theta^2):

sinโก(1.096)โ‰ˆ0.891\sin(1.096) \approx 0.891

Final Answers:

  1. sinโก2(ฯ€3)โ‰ˆ0.750\sin^2\left(\frac{\pi}{3}\right) \approx 0.750
  2. sinโก(ฮธ2)โ‰ˆ0.891\sin\left(\theta^2\right) \approx 0.891

Would you like any further explanation or have additional questions? Here are some related questions to explore:

  1. What is the exact value of sinโก(ฯ€3)\sin(\frac{\pi}{3}) without approximation?
  2. How does the graph of sinโก2(x)\sin^2(x) behave compared to sinโก(x2)\sin(x^2)?
  3. What is the relationship between sinโก(ฮธ2)\sin(\theta^2) and ฮธ2\theta^2 in terms of periodicity?
  4. Can you derive an approximation for sinโก(x)\sin(x) using a Taylor series expansion?
  5. How does squaring sinโก(ฮธ)\sin(\theta) affect the amplitude of the sine function?

Tip: Always be mindful of how functions behave at key angles like ฯ€6\frac{\pi}{6}, ฯ€4\frac{\pi}{4}, and ฯ€3\frac{\pi}{3}, as these yield neat, often rational results.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Sine Function
Square of Sine
Angle Squaring

Formulas

sin(๐œƒ) = โˆš3 / 2
sin^2(๐œƒ) = (sin(๐œƒ))^2
sin(๐œƒ^2) = sin(๐œƒ^2)

Theorems

Basic Trigonometric Identities

Suitable Grade Level

Grades 9-11