Finding $\sin(\frac{\theta}{2})$ When $\,\cos(\theta) = \frac{3}{5}$ In Quadrant I A Step-by-Step Solution

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In the realm of trigonometry, navigating trigonometric identities and quadrant rules is crucial for solving complex problems. This article delves into a specific problem: given that $,\cos(\theta) = \frac{3}{5}$ and $,\theta$ lies in Quadrant I, we aim to find the value of $,\sin(\frac{\theta}{2})$. This exploration requires a blend of trigonometric knowledge, including half-angle formulas and understanding quadrant-specific trigonometric values. Let's embark on this mathematical journey to unravel the solution step by step.

Understanding the Problem

At the heart of our problem lies the need to determine $,\sin(\frac{\theta}{2})$ knowing that $,\cos(\theta) = \frac{3}{5}$ and $,\theta$ is in the first quadrant. This involves employing trigonometric identities, particularly the half-angle formulas, which connect trigonometric functions of an angle to those of its half-angle. Before diving into the calculations, let's clarify the given information.

  • The Given Cosine Value: We know that $,\cos(\theta) = \frac{3}{5}$. This tells us the ratio of the adjacent side to the hypotenuse in a right-angled triangle where $,\theta$ is one of the acute angles.
  • Quadrant I: The fact that $,\theta$ is in Quadrant I is critical. In this quadrant, all trigonometric functions (sine, cosine, tangent, etc.) are positive. This helps us in choosing the correct sign when dealing with square roots in our calculations.
  • The Goal: Our objective is to find the value of $,\sin(\frac{\theta}{2})$ . This means we need to find the sine of half the angle $,\theta$. This is where the half-angle formula for sine comes into play.

Understanding these foundational elements sets the stage for applying the appropriate formulas and arriving at the solution. The half-angle formulas are powerful tools in trigonometry, allowing us to relate trigonometric functions of an angle to those of its half-angle or double angle. These formulas are derived from the fundamental trigonometric identities and are essential for solving a wide range of problems. In our case, the half-angle formula for sine is particularly relevant.

The Half-Angle Formula for Sine

The half-angle formula for sine is a cornerstone in solving our problem. It mathematically relates the sine of half an angle to the cosine of the full angle. This formula is expressed as:

sin⁡(θ2)=±1−cos⁡(θ)2\sin(\frac{\theta}{2}) = \pm \sqrt{\frac{1 - \cos(\theta)}{2}}

This formula is derived from the double-angle formula for cosine and is a powerful tool for finding the sine of half an angle when the cosine of the full angle is known. The "$,\pm$ " sign in front of the square root indicates that the result could be positive or negative, depending on the quadrant in which $,\frac{\theta}{2}$ lies. This is a crucial point to consider to ensure the correct sign is chosen for the final answer.

Applying the Formula

Now that we have the half-angle formula, let's apply it to our problem. We know that $,\cos(\theta) = \frac{3}{5}$, and we need to find $,\sin(\frac{\theta}{2})$. Substituting the given value into the formula, we get:

sin⁡(θ2)=±1−352\sin(\frac{\theta}{2}) = \pm \sqrt{\frac{1 - \frac{3}{5}}{2}}

This substitution is the first step in our calculation. Next, we need to simplify the expression under the square root. This involves basic arithmetic operations, including finding a common denominator and subtracting fractions. The goal is to reduce the expression to a simple fraction, which we can then take the square root of.

Simplifying the Expression

Let's simplify the expression under the square root:

sin⁡(θ2)=±55−352=±252\sin(\frac{\theta}{2}) = \pm \sqrt{\frac{\frac{5}{5} - \frac{3}{5}}{2}} = \pm \sqrt{\frac{\frac{2}{5}}{2}}

Here, we've rewritten 1 as $,\frac{5}{5}$ to have a common denominator with $,\frac{3}{5}$. Subtracting the fractions in the numerator, we get $,\frac{2}{5}$. Now, we have a fraction divided by 2, which we can simplify further:

sin⁡(θ2)=±25⋅12=±15\sin(\frac{\theta}{2}) = \pm \sqrt{\frac{2}{5} \cdot \frac{1}{2}} = \pm \sqrt{\frac{1}{5}}

We've simplified the expression by multiplying $,\frac{2}{5}$ by $,\frac{1}{2}$, which gives us $,\frac{1}{5}$. Now, we need to take the square root of $,\frac{1}{5}$. This involves taking the square root of both the numerator and the denominator separately.

Taking the Square Root

Taking the square root of $,\frac{1}{5}$ gives us:

sin⁡(θ2)=±15=±15\sin(\frac{\theta}{2}) = \pm \frac{\sqrt{1}}{\sqrt{5}} = \pm \frac{1}{\sqrt{5}}

We know that the square root of 1 is 1. Now we have $,\pm \frac{1}{\sqrt{5}}$. To rationalize the denominator, we multiply both the numerator and the denominator by $,\sqrt{5}$:

sin⁡(θ2)=±15⋅55=±55\sin(\frac{\theta}{2}) = \pm \frac{1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \pm \frac{\sqrt{5}}{5}

Now, we have a simplified expression for $,\sin(\frac{\theta}{2})$. However, we still need to determine the correct sign (positive or negative) based on the quadrant in which $,\frac{\theta}{2}$ lies. This is a crucial step to ensure the accuracy of our final answer.

Determining the Correct Sign

To determine the correct sign for $,\sin(\frac{\theta}{2})$, we need to consider the quadrant in which $,\frac{\theta}{2}$ lies. We know that $,\theta$ is in Quadrant I, which means that $0^\circ < \theta < 90^\circ$. Dividing this inequality by 2, we get:

0∘<θ2<45∘0^\circ < \frac{\theta}{2} < 45^\circ

This tells us that $,\frac{\theta}{2}$ also lies in Quadrant I. In Quadrant I, all trigonometric functions, including sine, are positive. Therefore, we choose the positive sign for $,\sin(\frac{\theta}{2})$:

sin⁡(θ2)=55\sin(\frac{\theta}{2}) = \frac{\sqrt{5}}{5}

This step is crucial as it ensures that our final answer aligns with the quadrant rules of trigonometry. Choosing the correct sign is just as important as the numerical calculation itself. A mistake in the sign can lead to a completely different answer.

The Final Answer

After careful calculation and consideration of the quadrant rules, we have arrived at the final answer:

sin⁡(θ2)=55\sin(\frac{\theta}{2}) = \frac{\sqrt{5}}{5}

This is the value of the sine of half the angle $,\theta$ given that $,\cos(\theta) = \frac{3}{5}$ and $,\theta$ is in Quadrant I. This result is a testament to the power of trigonometric identities and the importance of understanding quadrant-specific rules.

Verification and Conclusion

To ensure the correctness of our answer, it's always a good practice to verify it. One way to do this is to check if the value we obtained makes sense in the context of the problem. Since $,\frac{\theta}{2}$ is in Quadrant I, its sine should be a positive value between 0 and 1. Our answer, $,\frac{\sqrt{5}}{5}$, approximately equals 0.447, which fits this criterion.

In conclusion, by applying the half-angle formula for sine and considering the quadrant in which the angle lies, we have successfully found the value of $,\sin(\frac{\theta}{2})$. This problem highlights the interconnectedness of trigonometric concepts and the importance of a systematic approach to problem-solving in mathematics. The journey from the initial problem statement to the final answer showcases the elegance and power of trigonometric principles.

Let's explore how to calculate $,\sin(\frac{\theta}{2})$ given that $,\cos(\theta) = \frac{3}{5}$ and $,\theta$ is in the first quadrant. This involves understanding trigonometric identities and the implications of quadrant positioning.

Problem Restatement: Finding $\sin(\frac{\theta}{2})$ Given $,\cos(\theta) = \frac{3}{5}$

Our core task is to determine the value of $,\sin(\frac{\theta}{2})$ under the conditions that $,\cos(\theta) = \frac{3}{5}$ and $,\theta$ resides in Quadrant I. This requires a firm grasp of trigonometric identities, especially the half-angle formulas, and the properties that trigonometric functions exhibit in different quadrants. We'll dissect this problem meticulously, laying out each step to ensure clarity and comprehension.

Key Information: Decoding the Givens

Before we delve into the mathematical manipulations, let's solidify our understanding of the given information. This groundwork is crucial for a correct and efficient solution.

  1. The Cosine Value: We are provided with $,\cos(\theta) = \frac{3}{5}$. Cosine, in a right-angled triangle context, represents the ratio of the adjacent side to the hypotenuse. This numerical value is a cornerstone for our calculations.
  2. Quadrant I Location: The stipulation that $,\theta$ is in Quadrant I is of paramount importance. In Quadrant I, all trigonometric functions (sine, cosine, tangent, and their reciprocals) are positive. This fact will dictate the sign of our final answer, especially when dealing with square roots.
  3. The Objective: Our mission is crystal clear – we need to compute the exact value of $,\sin(\frac{\theta}{2})$. This necessitates the use of the half-angle formula for sine, a powerful tool in our trigonometric arsenal.

By fully grasping these givens, we position ourselves for a strategic application of trigonometric principles. The half-angle formulas are derived from fundamental trigonometric identities and are instrumental in connecting trigonometric functions of an angle with those of its half or double angle. In our specific scenario, the half-angle formula for sine takes center stage.

Half-Angle Formula: The Key to Unlocking $\sin(\frac{\theta}{2})$

The half-angle formula for sine is the linchpin in solving our problem. It establishes a direct mathematical relationship between the sine of half an angle and the cosine of the original angle. The formula is elegantly expressed as:

sin⁡(θ2)=±1−cos⁡(θ)2\sin(\frac{\theta}{2}) = \pm \sqrt{\frac{1 - \cos(\theta)}{2}}

This formula stems from the double-angle formula for cosine and is a pivotal tool for determining the sine of half an angle when the cosine of the full angle is known. The "$,\pm$ " symbol preceding the square root is a crucial reminder that the result can be either positive or negative. The correct sign depends entirely on the quadrant in which $,\frac{\theta}{2}$ resides, a factor we must carefully consider.

Formula Application: Plugging in the Known Value

Armed with the half-angle formula, we now substitute the given value of $,\cos(\theta) = \frac{3}{5}$ into the equation:

sin⁡(θ2)=±1−352\sin(\frac{\theta}{2}) = \pm \sqrt{\frac{1 - \frac{3}{5}}{2}}

This substitution is our first concrete step toward a numerical solution. The next phase involves simplifying the expression nestled under the square root. This requires basic yet crucial arithmetic operations, including finding common denominators and performing fraction subtraction. Our aim is to distill the expression into a manageable fraction ready for the square root operation.

Expression Simplification: Taming the Square Root

Let's meticulously simplify the expression under the square root:

sin⁡(θ2)=±55−352=±252\sin(\frac{\theta}{2}) = \pm \sqrt{\frac{\frac{5}{5} - \frac{3}{5}}{2}} = \pm \sqrt{\frac{\frac{2}{5}}{2}}

Here, we've intelligently rewritten 1 as $,\frac{5}{5}$ to achieve a common denominator with $,\frac{3}{5}$. Subtracting the fractions in the numerator yields $,\frac{2}{5}$. We now face a fraction divided by 2, which we can further simplify:

sin⁡(θ2)=±25⋅12=±15\sin(\frac{\theta}{2}) = \pm \sqrt{\frac{2}{5} \cdot \frac{1}{2}} = \pm \sqrt{\frac{1}{5}}

We've streamlined the expression by multiplying $,\frac{2}{5}$ by $,\frac{1}{2}$, resulting in $,\frac{1}{5}$. The next step is to evaluate the square root of $,\frac{1}{5}$. This entails finding the square root of both the numerator and the denominator separately.

Square Root Evaluation: Unveiling the Radical

Taking the square root of $,\frac{1}{5}$ gives us:

sin⁡(θ2)=±15=±15\sin(\frac{\theta}{2}) = \pm \frac{\sqrt{1}}{\sqrt{5}} = \pm \frac{1}{\sqrt{5}}

We know that the square root of 1 is 1. Now we have $,\pm \frac{1}{\sqrt{5}}$. To rationalize the denominator—a standard practice in mathematical simplification—we multiply both the numerator and the denominator by $,\sqrt{5}$:

sin⁡(θ2)=±15⋅55=±55\sin(\frac{\theta}{2}) = \pm \frac{1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \pm \frac{\sqrt{5}}{5}

We now have a simplified expression for $,\sin(\frac{\theta}{2})$. However, we must still determine the correct sign (positive or negative) based on the quadrant in which $,\frac{\theta}{2}$ is located. This is a pivotal step to ensure the accuracy of our final result.

Sign Determination: Navigating the Quadrants

To pinpoint the correct sign for $,\sin(\frac{\theta}{2})$, we must ascertain the quadrant in which $,\frac{\theta}{2}$ lies. We are given that $,\theta$ is in Quadrant I, meaning $0^\circ < \theta < 90^\circ$. Dividing this inequality by 2, we obtain:

0∘<θ2<45∘0^\circ < \frac{\theta}{2} < 45^\circ

This crucial inequality reveals that $,\frac{\theta}{2}$ also resides in Quadrant I. In Quadrant I, all trigonometric functions, including sine, are strictly positive. Therefore, we confidently select the positive sign for $,\sin(\frac{\theta}{2})$:

sin⁡(θ2)=55\sin(\frac{\theta}{2}) = \frac{\sqrt{5}}{5}

This step is of paramount importance as it ensures that our final answer is consistent with the fundamental quadrant rules of trigonometry. Correct sign selection is as critical as the numerical computation itself. An error in sign can lead to a completely erroneous result.

The Solution: Unveiling $\sin(\frac{\theta}{2})$

Following a rigorous process of calculation and careful consideration of quadrant rules, we arrive at the definitive solution:

sin⁡(θ2)=55\sin(\frac{\theta}{2}) = \frac{\sqrt{5}}{5}

This represents the value of the sine of half the angle $,\theta$ given the initial conditions: $,\cos(\theta) = \frac{3}{5}$ and $,\theta$ in Quadrant I. This solution underscores the power of trigonometric identities and the necessity of understanding quadrant-specific behavior.

Solution Verification and Conclusion

To bolster our confidence in the correctness of our answer, a verification step is highly recommended. One effective method is to assess the reasonableness of the obtained value within the problem's context. Given that $,\frac{\theta}{2}$ is in Quadrant I, its sine should be a positive value strictly between 0 and 1. Our result, $,\frac{\sqrt{5}}{5}$, is approximately 0.447, which comfortably satisfies this criterion.

In summary, we have successfully determined the value of $,\sin(\frac{\theta}{2})$ by skillfully applying the half-angle formula for sine and meticulously accounting for the relevant quadrant rules. This problem serves as a compelling illustration of the interconnected nature of trigonometric concepts and the value of a systematic, step-by-step approach to mathematical problem-solving. The journey from the problem's inception to the final solution showcases the inherent elegance and utility of trigonometric principles.

Step-by-Step Guide to Finding $\sin(\frac{\theta}{2})$ When $,\cos(\theta) = \frac{3}{5}$ in Quadrant I

Are you grappling with the challenge of finding $,\sin(\frac{\theta}{2})$ when $,\cos(\theta) = \frac{3}{5}$ and $,\theta$ resides in the first quadrant? This article provides a comprehensive, step-by-step guide to navigate this trigonometric problem with confidence. We'll break down the process into manageable steps, ensuring a clear understanding of the concepts and calculations involved.

1. Understand the Basics: Givens and Goal

Before diving into calculations, let's solidify our understanding of the given information. This groundwork is crucial for a correct and efficient solution. We are given:

  • \,\cos(\theta) = \frac{3}{5}$: This tells us the ratio of the adjacent side to the hypotenuse in a right-angled triangle.

  • \,\theta$ is in Quadrant I: This means all trigonometric functions are positive.

Our goal is to find:

  • \,\sin(\frac{\theta}{2})$: The sine of half the angle $\,\theta$.

This initial assessment sets the stage for applying the appropriate trigonometric principles. We'll be leveraging the half-angle formula for sine, a powerful tool that connects the sine of half an angle to the cosine of the full angle.

2. Apply the Half-Angle Formula for Sine

The cornerstone of our solution is the half-angle formula for sine, which states:

sin⁡(θ2)=±1−cos⁡(θ)2\sin(\frac{\theta}{2}) = \pm \sqrt{\frac{1 - \cos(\theta)}{2}}

This formula is derived from the double-angle formula for cosine and is essential for problems involving half angles. Notice the "$,\pm$ " sign, which indicates that we need to determine the correct sign based on the quadrant of $,\frac{\theta}{2}$.

Substitute the given value of $,\cos(\theta) = \frac{3}{5}$ into the formula:

sin⁡(θ2)=±1−352\sin(\frac{\theta}{2}) = \pm \sqrt{\frac{1 - \frac{3}{5}}{2}}

This substitution is the first step in our calculation. Next, we'll simplify the expression under the square root, a process that involves basic arithmetic operations.

3. Simplify the Expression

Simplify the expression under the square root step-by-step:

sin⁡(θ2)=±55−352=±252\sin(\frac{\theta}{2}) = \pm \sqrt{\frac{\frac{5}{5} - \frac{3}{5}}{2}} = \pm \sqrt{\frac{\frac{2}{5}}{2}}

Here, we rewrote 1 as $,\frac{5}{5}$ to have a common denominator. Now, subtract the fractions:

sin⁡(θ2)=±25÷2=±15\sin(\frac{\theta}{2}) = \pm \sqrt{\frac{2}{5} \div 2} = \pm \sqrt{\frac{1}{5}}

We simplified the fraction within the square root. Now, we proceed to take the square root of the simplified fraction.

4. Take the Square Root

Evaluate the square root:

sin⁡(θ2)=±15=±15\sin(\frac{\theta}{2}) = \pm \frac{\sqrt{1}}{\sqrt{5}} = \pm \frac{1}{\sqrt{5}}

We know that the square root of 1 is 1. To rationalize the denominator (eliminate the square root from the denominator), multiply both the numerator and the denominator by $,\sqrt{5}$:

sin⁡(θ2)=±15⋅55=±55\sin(\frac{\theta}{2}) = \pm \frac{1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \pm \frac{\sqrt{5}}{5}

Now we have a simplified expression, but we still need to determine the correct sign based on the quadrant in which $,\frac{\theta}{2}$ lies.

5. Determine the Sign Based on Quadrant

To determine the sign, we need to find the quadrant of $,\frac{\theta}{2}$. Since $,\theta$ is in Quadrant I, we know that $0^\circ < \theta < 90^\circ$. Divide this inequality by 2:

0∘<θ2<45∘0^\circ < \frac{\theta}{2} < 45^\circ

This tells us that $,\frac{\theta}{2}$ is also in Quadrant I. In Quadrant I, sine is positive. Therefore, we choose the positive sign:

sin⁡(θ2)=55\sin(\frac{\theta}{2}) = \frac{\sqrt{5}}{5}

This step is crucial for ensuring the accuracy of our final answer. The quadrant in which the angle lies dictates the sign of the trigonometric function.

6. State the Final Answer

After careful calculation and consideration of the quadrant rules, we have found the value of $,\sin(\frac{\theta}{2})$:

sin⁡(θ2)=55\sin(\frac{\theta}{2}) = \frac{\sqrt{5}}{5}

This is the sine of half the angle $,\theta$ given the initial conditions. To verify our answer, we can check if the value makes sense within the context of the problem.

7. Verify the Solution

To verify our solution, we can check if the value we obtained is reasonable. Since $,\frac{\theta}{2}$ is in Quadrant I, its sine should be a positive value between 0 and 1. Our answer, $,\frac{\sqrt{5}}{5}$, is approximately 0.447, which falls within this range.

Conclusion

By following these step-by-step instructions, you can confidently find $,\sin(\frac{\theta}{2})$ when given $,\cos(\theta)$ and the quadrant of $,\theta$. This problem highlights the importance of understanding trigonometric identities and quadrant rules. With practice, you can master these concepts and tackle a wide range of trigonometric problems. Remember, the key is to break down the problem into manageable steps and apply the appropriate formulas and principles.