Simplifying Trigonometric Expressions A Step By Step Guide

Trigonometry, a cornerstone of mathematics, involves the study of relationships between angles and sides of triangles. It extends to a broader field analyzing periodic phenomena, making it indispensable in various scientific and engineering disciplines. Mastering trigonometric identities and simplification techniques is crucial for problem-solving in mathematics and its applications. This article delves into several trigonometric problems, providing step-by-step solutions and explanations to enhance understanding and proficiency.

I. Simplifying ${\sin 78^\circ - \sin 18^\circ + \sin 30^\circ - \sin 42^\circ}$

Trigonometric simplification often involves using identities to rewrite expressions into simpler forms. In this case, we aim to simplify the expression ${\sin 78^\circ - \sin 18^\circ + \sin 30^\circ - \sin 42^\circ}$. The key lies in strategically applying trigonometric identities to combine terms and simplify the expression. We can begin by grouping terms and using the sine subtraction formula, which states that ${\sin A - \sin B = 2 \cos(\frac{A + B}{2}) \sin(\frac{A - B}{2})}$. By applying this formula, we can transform the differences of sine functions into products, which are often easier to manage. Let’s start by grouping the first two terms and the last term with ${\sin 30^\circ}$.

Applying the sine subtraction formula to ${\sin 78^\circ - \sin 18^\circ}$, we get:

${\sin 78^\circ - \sin 18^\circ = 2 \cos(\frac{78^\circ + 18^\circ}{2}) \sin(\frac{78^\circ - 18^\circ}{2}) = 2 \cos(48^\circ) \sin(30^\circ)}$

Since ${\sin 30^\circ = \frac{1}{2}}$, this simplifies to:

${2 \cos(48^\circ) \sin(30^\circ) = 2 \cos(48^\circ) \cdot \frac{1}{2} = \cos(48^\circ)}$

Now, let's consider the term ${\sin 30^\circ - \sin 42^\circ}$. Again, applying the sine subtraction formula:

${\sin 30^\circ - \sin 42^\circ = 2 \cos(\frac{30^\circ + 42^\circ}{2}) \sin(\frac{30^\circ - 42^\circ}{2}) = 2 \cos(36^\circ) \sin(-6^\circ)}$

Since ${\sin(-x) = -\sin(x)}$, this becomes:

${2 \cos(36^\circ) \sin(-6^\circ) = -2 \cos(36^\circ) \sin(6^\circ)}$

Now, we combine the simplified terms:

${\sin 78^\circ - \sin 18^\circ + \sin 30^\circ - \sin 42^\circ = \cos(48^\circ) - 2 \cos(36^\circ) \sin(6^\circ)}$

We can rewrite ${\cos(48^\circ)}$ as ${\sin(42^\circ)}$ using the identity ${\cos(x) = \sin(90^\circ - x)}$. So, the expression becomes:

${\sin(42^\circ) - 2 \cos(36^\circ) \sin(6^\circ)}$

To further simplify, we can use the product-to-sum identity, which states that ${2 \sin A \cos B = \sin(A + B) + \sin(A - B)}$. Applying this to the second term:

${2 \cos(36^\circ) \sin(6^\circ) = \sin(36^\circ + 6^\circ) + \sin(6^\circ - 36^\circ) = \sin(42^\circ) + \sin(-30^\circ)}$

Since ${\sin(-30^\circ) = -\sin(30^\circ) = -\frac{1}{2}}$, we have:

${\sin(42^\circ) + \sin(-30^\circ) = \sin(42^\circ) - \frac{1}{2}}$

Substituting this back into the original expression:

${\sin(42^\circ) - (\sin(42^\circ) - \frac{1}{2}) = \sin(42^\circ) - \sin(42^\circ) + \frac{1}{2} = \frac{1}{2}}$

Thus, ${\sin 78^\circ - \sin 18^\circ + \sin 30^\circ - \sin 42^\circ = \frac{1}{2}}$.

II. Simplifying ${\sec 20^\circ \sec 40^\circ \sec 80^\circ}$

The simplification of trigonometric expressions involving secant functions often requires converting them to cosine functions, as cosine functions have well-known product-to-sum identities. To simplify ${\sec 20^\circ \sec 40^\circ \sec 80^\circ}$, we first rewrite the secant functions in terms of cosine:

${\sec 20^\circ \sec 40^\circ \sec 80^\circ = \frac{1}{\cos 20^\circ} \cdot \frac{1}{\cos 40^\circ} \cdot \frac{1}{\cos 80^\circ} = \frac{1}{\cos 20^\circ \cos 40^\circ \cos 80^\circ}}$

Now, we focus on simplifying the denominator, ${\cos 20^\circ \cos 40^\circ \cos 80^\circ}$. A common strategy for such products is to multiply and divide by a suitable sine function to utilize the double-angle identity. In this case, we multiply and divide by ${\sin 20^\circ}$:

${\frac{\cos 20^\circ \cos 40^\circ \cos 80^\circ}{1} = \frac{\sin 20^\circ \cos 20^\circ \cos 40^\circ \cos 80^\circ}{\sin 20^\circ}}$

Using the identity ${2 \sin x \cos x = \sin 2x}$, we can simplify ${\sin 20^\circ \cos 20^\circ}$ to ${\frac{1}{2} \sin 40^\circ}$:

${\frac{\frac{1}{2} \sin 40^\circ \cos 40^\circ \cos 80^\circ}{\sin 20^\circ}}$

Next, we apply the same identity to ${\sin 40^\circ \cos 40^\circ}$, which gives us ${\frac{1}{2} \sin 80^\circ}$:

${\frac{\frac{1}{2} \cdot \frac{1}{2} \sin 80^\circ \cos 80^\circ}{\sin 20^\circ} = \frac{\frac{1}{4} \sin 80^\circ \cos 80^\circ}{\sin 20^\circ}}$

Applying the double-angle identity again, we simplify ${\sin 80^\circ \cos 80^\circ}$ to ${\frac{1}{2} \sin 160^\circ}$:

${\frac{\frac{1}{4} \cdot \frac{1}{2} \sin 160^\circ}{\sin 20^\circ} = \frac{\frac{1}{8} \sin 160^\circ}{\sin 20^\circ}}$

Since ${\sin(180^\circ - x) = \sin x}$, we have ${\sin 160^\circ = \sin(180^\circ - 20^\circ) = \sin 20^\circ}$. Thus, the expression simplifies to:

${\frac{\frac{1}{8} \sin 20^\circ}{\sin 20^\circ} = \frac{1}{8}}$

Now, we substitute this back into the original expression:

${\sec 20^\circ \sec 40^\circ \sec 80^\circ = \frac{1}{\cos 20^\circ \cos 40^\circ \cos 80^\circ} = \frac{1}{\frac{1}{8}} = 8}$

Therefore, ${\sec 20^\circ \sec 40^\circ \sec 80^\circ = 8}$.

III. Simplifying ${\sqrt{3} \cot 20^\circ \cot 40^\circ \cot 80^\circ}$

Simplifying expressions involving cotangent functions often benefits from converting them to tangent functions or sine and cosine functions, which allows for the application of various trigonometric identities. To simplify ${\sqrt{3} \cot 20^\circ \cot 40^\circ \cot 80^\circ}$, we first rewrite the cotangent functions in terms of tangent:

${\sqrt{3} \cot 20^\circ \cot 40^\circ \cot 80^\circ = \sqrt{3} \cdot \frac{1}{\tan 20^\circ} \cdot \frac{1}{\tan 40^\circ} \cdot \frac{1}{\tan 80^\circ} = \frac{\sqrt{3}}{\tan 20^\circ \tan 40^\circ \tan 80^\circ}}$

Now, we focus on simplifying the denominator, ${\tan 20^\circ \tan 40^\circ \tan 80^\circ}$. We can rewrite the tangent functions in terms of sine and cosine:

${\tan 20^\circ \tan 40^\circ \tan 80^\circ = \frac{\sin 20^\circ}{\cos 20^\circ} \cdot \frac{\sin 40^\circ}{\cos 40^\circ} \cdot \frac{\sin 80^\circ}{\cos 80^\circ}}$

To simplify this product, we can use the identity ${\sin x = \cos(90^\circ - x)}$ to rewrite ${\sin 80^\circ}$ as ${\cos 10^\circ}$ and ${\sin 40^\circ}$ as ${\cos 50^\circ}$. However, a more direct approach involves using the triple angle formula for tangent, but in a reverse manner. Let's consider the identity:

${\tan(3x) = \frac{3 \tan x - \tan^3 x}{1 - 3 \tan^2 x}}$

This identity doesn't directly help us simplify the product, so let's revert to using the sine and cosine representation and multiply the terms:

${\frac{\sin 20^\circ \sin 40^\circ \sin 80^\circ}{\cos 20^\circ \cos 40^\circ \cos 80^\circ}}$

We already know from the previous simplification that ${\cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{1}{8}}$. Now, let's simplify the numerator, ${\sin 20^\circ \sin 40^\circ \sin 80^\circ}$. We can use the product-to-sum identities:

First, apply the identity ${2 \sin A \sin B = \cos(A - B) - \cos(A + B)}$ to ${\sin 20^\circ \sin 40^\circ}$:

${2 \sin 20^\circ \sin 40^\circ = \cos(40^\circ - 20^\circ) - \cos(40^\circ + 20^\circ) = \cos 20^\circ - \cos 60^\circ}$

Since ${\cos 60^\circ = \frac{1}{2}}$, we have:

${\sin 20^\circ \sin 40^\circ = \frac{1}{2} (\cos 20^\circ - \frac{1}{2})}$

Now, multiply by ${\sin 80^\circ}$:

${\sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{1}{2} (\cos 20^\circ - \frac{1}{2}) \sin 80^\circ = \frac{1}{2} \cos 20^\circ \sin 80^\circ - \frac{1}{4} \sin 80^\circ}$

Using the product-to-sum identity ${2 \sin A \cos B = \sin(A + B) + \sin(A - B)}$:

${2 \cos 20^\circ \sin 80^\circ = \sin(80^\circ + 20^\circ) + \sin(80^\circ - 20^\circ) = \sin 100^\circ + \sin 60^\circ}$

Since ${\sin 100^\circ = \sin(180^\circ - 80^\circ) = \sin 80^\circ}$ and ${\sin 60^\circ = \frac{\sqrt{3}}{2}}$, we get:

${\cos 20^\circ \sin 80^\circ = \frac{1}{2} (\sin 80^\circ + \frac{\sqrt{3}}{2})}$

Substitute this back into the expression:

${\sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{1}{2} [\frac{1}{2} (\sin 80^\circ + \frac{\sqrt{3}}{2})] - \frac{1}{4} \sin 80^\circ}$

${= \frac{1}{4} \sin 80^\circ + \frac{\sqrt{3}}{8} - \frac{1}{4} \sin 80^\circ = \frac{\sqrt{3}}{8}}$

Now we have ${\tan 20^\circ \tan 40^\circ \tan 80^\circ = \frac{\sin 20^\circ \sin 40^\circ \sin 80^\circ}{\cos 20^\circ \cos 40^\circ \cos 80^\circ} = \frac{\frac{\sqrt{3}}{8}}{\frac{1}{8}} = \sqrt{3}}$.

Substitute this back into the original expression:

${\sqrt{3} \cot 20^\circ \cot 40^\circ \cot 80^\circ = \frac{\sqrt{3}}{\tan 20^\circ \tan 40^\circ \tan 80^\circ} = \frac{\sqrt{3}}{\sqrt{3}} = 1}$

Thus, ${\sqrt{3} \cot 20^\circ \cot 40^\circ \cot 80^\circ = 1}$.

IV. Simplifying ${\frac{1}{2 \sin 10^\circ} - 2 \sin 70^\circ}$

To simplify ${\frac{1}{2 \sin 10^\circ} - 2 \sin 70^\circ}$, we will use trigonometric identities to combine and reduce the expression. Our goal is to express the given expression in a simpler, more manageable form. First, let’s rewrite the expression:

${\frac{1}{2 \sin 10^\circ} - 2 \sin 70^\circ = \frac{1 - 4 \sin 10^\circ \sin 70^\circ}{2 \sin 10^\circ}}$

Now, we focus on simplifying the numerator. We can use the product-to-sum identity ${2 \sin A \sin B = \cos(A - B) - \cos(A + B)}$:

${4 \sin 10^\circ \sin 70^\circ = 2(2 \sin 10^\circ \sin 70^\circ) = 2[\cos(70^\circ - 10^\circ) - \cos(70^\circ + 10^\circ)]}$

${= 2(\cos 60^\circ - \cos 80^\circ)}$

Since ${\cos 60^\circ = \frac{1}{2}}$, we have:

${2(\cos 60^\circ - \cos 80^\circ) = 2(\frac{1}{2} - \cos 80^\circ) = 1 - 2 \cos 80^\circ}$

Substitute this back into the numerator:

${1 - 4 \sin 10^\circ \sin 70^\circ = 1 - (1 - 2 \cos 80^\circ) = 2 \cos 80^\circ}$

So, the expression becomes:

${\frac{2 \cos 80^\circ}{2 \sin 10^\circ} = \frac{\cos 80^\circ}{\sin 10^\circ}}$

Using the identity ${\sin x = \cos(90^\circ - x)}$, we can rewrite ${\sin 10^\circ}$ as ${\cos(90^\circ - 10^\circ) = \cos 80^\circ}$:

${\frac{\cos 80^\circ}{\sin 10^\circ} = \frac{\cos 80^\circ}{\cos 80^\circ} = 1}$

Thus, ${\frac{1}{2 \sin 10^\circ} - 2 \sin 70^\circ = 1}$.

Conclusion

In summary, this article demonstrated the simplification of various trigonometric expressions using standard trigonometric identities and techniques. By strategically applying these identities, we were able to reduce complex expressions to simpler forms. Mastering these techniques is essential for solving a wide range of problems in mathematics, physics, and engineering. The examples provided illustrate the importance of recognizing patterns, applying appropriate identities, and simplifying step-by-step to arrive at the solution. Through practice and a solid understanding of fundamental identities, one can develop the proficiency needed to tackle even more challenging trigonometric problems. Remember, the key to success in trigonometry lies in understanding the relationships between trigonometric functions and their applications.