Solving (sin Φ - Cos Φ) / (sin Φ + Cos Φ) Given Cot Φ = √3

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In this article, we will explore how to solve trigonometric expressions when given the value of the cotangent of an angle. Specifically, we'll tackle the problem: If $\cot \phi = \\sqrt{3}$, find the value of $\frac{\sin \phi - \cos \phi}{\sin \phi + \cos \phi}$. This problem involves understanding trigonometric identities, the relationship between trigonometric functions, and algebraic manipulation. We'll break down the solution step-by-step, making it easy to follow and understand.

Understanding the Basics of Trigonometric Functions

Before diving into the problem, let's quickly recap the basic trigonometric functions and their relationships. The primary trigonometric functions are sine (sin), cosine (cos), and tangent (tan). These functions relate the angles of a right-angled triangle to the ratios of its sides. The cotangent (cot) function is the reciprocal of the tangent function. Understanding these relationships is crucial for solving trigonometric problems.

  • Sine (sin θ) = Opposite / Hypotenuse
  • Cosine (cos θ) = Adjacent / Hypotenuse
  • Tangent (tan θ) = Opposite / Adjacent
  • Cotangent (cot θ) = Adjacent / Opposite = 1 / tan θ

Additionally, it's important to remember the fundamental trigonometric identity:

sin2theta+cos2theta=1\\sin^2 \\theta + \\cos^2 \\theta = 1

This identity is a cornerstone in trigonometry and is frequently used in simplifying and solving trigonometric expressions. In our problem, we are given that $\cot \phi = \\sqrt{3}$. This information, combined with our knowledge of trigonometric relationships, will help us find the values of $\sin \phi$ and $\cos \phi$, which are essential for solving the main expression.

Step-by-Step Solution

1. Find the angle $\phi$

We are given that $\cot \phi = \\sqrt{3}$. We know that cotangent is the reciprocal of tangent, so we can write:

tanphi=frac1cotphi=frac1sqrt3\\tan \\phi = \\frac{1}{\\cot \\phi} = \\frac{1}{\\\\sqrt{3}}

To rationalize the denominator, we multiply the numerator and denominator by $\\sqrt{3}$:

tanphi=frac1sqrt3timesfracsqrt3sqrt3=fracsqrt33\\tan \\phi = \\frac{1}{\\\\sqrt{3}} \\times \\frac{\\\\sqrt{3}}{\\\\sqrt{3}} = \\frac{\\\\sqrt{3}}{3}

Now, we need to find the angle $\phi$ whose tangent is $\frac{\\sqrt{3}}{3}$. We know from the standard trigonometric values that:

\\tan 30^\\circ = \\tan \\frac{\\pi}{6} = \\frac{\\\\sqrt{3}}{3}

Therefore, $\phi = 30^\circ$ or $\frac{\pi}{6}$ radians. It's important to consider the range of possible solutions, but for simplicity, we'll focus on the principal value here. However, for a more comprehensive understanding, one should acknowledge that tangent (and cotangent) functions have a period of $\\pi$, meaning there are multiple angles with the same tangent value. In this case, we are considering the simplest solution for this problem.

2. Find $\sin \phi$ and $\cos \phi$

Now that we have found $\phi = 30^\circ$, we can find the values of $\sin \phi$ and $\cos \phi$ using the standard trigonometric values:

\\sin \\phi = \\sin 30^\\circ = \\sin \\frac{\\pi}{6} = \\frac{1}{2}

\\cos \\phi = \\cos 30^\\circ = \\cos \\frac{\\pi}{6} = \\frac{\\\\sqrt{3}}{2}

These values are fundamental and should be memorized or quickly derivable from the unit circle or special right triangles (30-60-90 triangles).

3. Substitute the values into the expression

We are asked to find the value of:

fracsinphi−cosphisinphi+cosphi\\frac{\\sin \\phi - \\cos \\phi}{\\sin \\phi + \\cos \\phi}

Now, we substitute the values of $\sin \phi$ and $\cos \phi$ we found in the previous step:

fracfrac12−fracsqrt32frac12+fracsqrt32\\frac{\\frac{1}{2} - \\frac{\\\\sqrt{3}}{2}}{\\frac{1}{2} + \\frac{\\\\sqrt{3}}{2}}

4. Simplify the expression

To simplify this expression, we first find a common denominator in both the numerator and the denominator, which is already the case here (both have a denominator of 2):

fracfrac1−sqrt32frac1+sqrt32\\frac{\\frac{1 - \\\\sqrt{3}}{2}}{\\frac{1 + \\\\sqrt{3}}{2}}

Now, we can simplify by multiplying both the numerator and the denominator by 2:

frac1−sqrt31+sqrt3\\frac{1 - \\\\sqrt{3}}{1 + \\\\sqrt{3}}

To further simplify, we can rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator, which is $(1 - \\sqrt{3})$:

frac1−sqrt31+sqrt3timesfrac1−sqrt31−sqrt3=frac(1−sqrt3)2(1+sqrt3)(1−sqrt3)\\frac{1 - \\\\sqrt{3}}{1 + \\\\sqrt{3}} \\times \\frac{1 - \\\\sqrt{3}}{1 - \\\\sqrt{3}} = \\frac{(1 - \\\\sqrt{3})^2}{(1 + \\\\sqrt{3})(1 - \\\\sqrt{3})}

Expanding the numerator and denominator:

Numerator: $(1 - \\sqrt{3})^2 = 1 - 2\\sqrt{3} + 3 = 4 - 2\\sqrt{3}$

Denominator: $(1 + \\sqrt{3})(1 - \\sqrt{3}) = 1 - 3 = -2$

So, the expression becomes:

frac4−2sqrt3−2\\frac{4 - 2\\\\sqrt{3}}{-2}

Now, divide both terms in the numerator by -2:

frac4−2−frac2sqrt3−2=−2+sqrt3\\frac{4}{-2} - \\frac{2\\\\sqrt{3}}{-2} = -2 + \\\\sqrt{3}

Thus, the value of the expression is:

−2+sqrt3-2 + \\\\sqrt{3}

Alternative Method: Dividing Numerator and Denominator by Cos $\phi$

Another approach to solving this problem involves dividing both the numerator and the denominator of the original expression by $\cos \phi$. This can simplify the expression by introducing the tangent function, which is related to the given cotangent value.

Starting with the expression:

fracsinphi−cosphisinphi+cosphi\\frac{\\sin \\phi - \\cos \\phi}{\\sin \\phi + \\cos \\phi}

Divide both the numerator and the denominator by $\cos \phi$:

fracfracsinphicosphi−fraccosphicosphifracsinphicosphi+fraccosphicosphi\\frac{\\frac{\\sin \\phi}{\\cos \\phi} - \\frac{\\cos \\phi}{\\cos \\phi}}{\\frac{\\sin \\phi}{\\cos \\phi} + \\frac{\\cos \\phi}{\\cos \\phi}}

This simplifies to:

fractanphi−1tanphi+1\\frac{\\tan \\phi - 1}{\\tan \\phi + 1}

We already found that $\tan \phi = \frac{1}{\\sqrt{3}}$, so we substitute this value into the expression:

fracfrac1sqrt3−1frac1sqrt3+1\\frac{\\frac{1}{\\\\sqrt{3}} - 1}{\\frac{1}{\\\\sqrt{3}} + 1}

To simplify, multiply both the numerator and the denominator by $\\sqrt{3}$:

frac1−sqrt31+sqrt3\\frac{1 - \\\\sqrt{3}}{1 + \\\\sqrt{3}}

This is the same expression we obtained in the previous method, and we can simplify it as before by rationalizing the denominator:

frac1−sqrt31+sqrt3timesfrac1−sqrt31−sqrt3=frac(1−sqrt3)21−3=frac4−2sqrt3−2=−2+sqrt3\\frac{1 - \\\\sqrt{3}}{1 + \\\\sqrt{3}} \\times \\frac{1 - \\\\sqrt{3}}{1 - \\\\sqrt{3}} = \\frac{(1 - \\\\sqrt{3})^2}{1 - 3} = \\frac{4 - 2\\\\sqrt{3}}{-2} = -2 + \\\\sqrt{3}

This alternative method confirms our previous result, providing a different perspective on solving the problem.

Conclusion

In conclusion, if $\cot \phi = \\sqrt{3}$, the value of $\frac{\sin \phi - \cos \phi}{\sin \phi + \cos \phi}$ is $-2 + \\sqrt{3}$. We arrived at this solution by first finding the angle $\phi$ using the given cotangent value, then determining the corresponding sine and cosine values, and finally substituting these values into the expression and simplifying. We also demonstrated an alternative method involving dividing by cosine to simplify the expression using tangent. Understanding these steps and methods is essential for mastering trigonometric problem-solving. This problem highlights the importance of knowing trigonometric identities, standard values, and algebraic manipulation techniques in solving trigonometric expressions.

By breaking down the problem into smaller, manageable steps, we can tackle even complex trigonometric questions with confidence. The key is to have a solid foundation in the basic trigonometric functions and their relationships, as well as the ability to apply algebraic techniques to simplify expressions. Remember to always double-check your work and consider alternative methods to verify your solution.

This article provides a detailed explanation of how to solve this specific trigonometric problem. By understanding the underlying principles and applying them systematically, you can tackle a wide range of similar problems. Always focus on understanding the concepts rather than just memorizing steps, and practice regularly to improve your problem-solving skills. Mastering trigonometry is crucial for various fields, including mathematics, physics, and engineering, making it a valuable skill to develop.