Finding Sin(θ) Given Cos(θ) In Quadrant I

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In this comprehensive guide, we will delve into the process of determining the sine of an angle (θ{\theta}) when its cosine is known and the angle lies within the first quadrant. This is a fundamental concept in trigonometry, with applications spanning various fields such as physics, engineering, and computer graphics.

Understanding the Problem

We are given the equation:

cos(θ)=115\cos(\theta) = \frac{\sqrt{11}}{5}

And we know that $\theta$ is an angle in Quadrant I. Our goal is to find the value of $\sin(\theta)$.

Understanding the Fundamentals: Before we dive into the solution, let's refresh some key trigonometric concepts. The unit circle is a circle with a radius of 1 centered at the origin of a coordinate plane. For any angle $\theta$, the point where the terminal side of the angle intersects the unit circle has coordinates (cos( heta), sin( heta)).

Key Trigonometric Identity: The most important identity we'll use is the Pythagorean identity:

sin2(θ)+cos2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1

This identity is derived from the Pythagorean theorem applied to the right triangle formed by the point on the unit circle, the origin, and the projection of the point onto the x-axis. In essence, the identity states that the sum of the squares of the sine and cosine of any angle is always equal to 1.

Quadrants and Signs: The coordinate plane is divided into four quadrants, each with specific sign conventions for trigonometric functions. In Quadrant I, both the x and y coordinates are positive. Therefore, both the cosine and sine of an angle in Quadrant I are positive. This is crucial for our problem, as it tells us that the sine value we are looking for will be positive.

Why is this important? Understanding the quadrant in which the angle lies is crucial because it dictates the signs of trigonometric functions. In Quadrant I, sine, cosine, and tangent are all positive. In Quadrant II, sine is positive, while cosine and tangent are negative. In Quadrant III, tangent is positive, while sine and cosine are negative. And in Quadrant IV, cosine is positive, while sine and tangent are negative. Ignoring the quadrant information can lead to incorrect solutions.

Step-by-Step Solution

Now, let's solve the problem step by step:

Step 1: Utilize the Pythagorean Identity: We know that $\cos(\theta) = \frac{\sqrt{11}}{5}$. We can substitute this value into the Pythagorean identity:

sin2(θ)+(115)2=1\sin^2(\theta) + \left(\frac{\sqrt{11}}{5}\right)^2 = 1

This substitution allows us to relate the sine and cosine of the angle, which is the key to finding the unknown sine value.

Step 2: Simplify the Equation: Next, we simplify the equation by squaring the cosine term:

sin2(θ)+1125=1\sin^2(\theta) + \frac{11}{25} = 1

This simplification makes the equation easier to manipulate and isolate the $\sin^2(\theta)$ term.

Step 3: Isolate $\sin^2(\theta)$: To isolate $\sin^2(\theta)$, we subtract $\frac{11}{25}$ from both sides of the equation:

sin2(θ)=11125\sin^2(\theta) = 1 - \frac{11}{25}

This step brings us closer to solving for the sine value by isolating the squared sine term.

Step 4: Find a Common Denominator and Subtract: We need a common denominator to subtract the fractions. The common denominator for 1 and $\frac{11}{25}$ is 25. So, we rewrite 1 as $\frac{25}{25}$:

sin2(θ)=25251125\sin^2(\theta) = \frac{25}{25} - \frac{11}{25}

Now, we can subtract the fractions:

sin2(θ)=1425\sin^2(\theta) = \frac{14}{25}

This step gives us the value of $\sin^2(\theta)$.

Step 5: Take the Square Root: To find $\sin(\theta)$, we take the square root of both sides of the equation:

sin(θ)=±1425\sin(\theta) = \pm\sqrt{\frac{14}{25}}

Remember that taking the square root yields both positive and negative solutions. This is where the quadrant information becomes crucial.

Step 6: Simplify the Square Root: We can simplify the square root by taking the square root of the numerator and the denominator separately:

sin(θ)=±1425\sin(\theta) = \pm\frac{\sqrt{14}}{\sqrt{25}}

sin(θ)=±145\sin(\theta) = \pm\frac{\sqrt{14}}{5}

This simplification makes the result more readable and easier to work with.

Step 7: Determine the Sign: Since $\theta$ is in Quadrant I, we know that $\sin(\theta)$ is positive. Therefore, we choose the positive solution:

sin(θ)=145\sin(\theta) = \frac{\sqrt{14}}{5}

This final step gives us the correct value for $\sin(\theta)$, considering the quadrant in which the angle lies.

The Answer

Therefore, if $\cos(\theta) = \frac{\sqrt{11}}{5}$ and $\theta$ is in Quadrant I, then:

sin(θ)=145\sin(\theta) = \frac{\sqrt{14}}{5}

This is the solution to the problem. We have successfully found the sine of the angle using the given information and the Pythagorean identity.

Key Takeaways

  • The Pythagorean identity (sin2(θ)+cos2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1) is a fundamental tool for relating sine and cosine.
  • Understanding quadrants is essential for determining the correct sign of trigonometric functions.
  • Simplifying radicals makes the answer easier to understand and use.

Practice Problems

To solidify your understanding, try solving these similar problems:

  1. If $\cos(\theta) = \frac{3}{5}$ and $\theta$ is in Quadrant I, find $\sin(\theta)$.
  2. If $\sin(\theta) = \frac{1}{2}$ and $\theta$ is in Quadrant II, find $\cos(\theta)$.
  3. If $\cos(\theta) = -\frac{2}{3}$ and $\theta$ is in Quadrant III, find $\sin(\theta)$.

By working through these problems, you'll reinforce your understanding of the concepts and techniques discussed in this guide. Remember to pay close attention to the quadrant information and use the Pythagorean identity to relate sine and cosine.

Conclusion

Finding the sine of an angle when its cosine is known and the angle's quadrant is specified is a common trigonometric problem. By understanding the Pythagorean identity, quadrant rules, and following a step-by-step approach, you can confidently solve these problems. This skill is crucial for further studies in trigonometry and its applications in various fields. Remember to practice regularly and apply these concepts to different problems to master them. Happy calculating!