Hey guys! Today, we're diving into the exciting world of trigonometry, and we're going to tackle a problem that might seem a bit daunting at first glance. But don't worry, we'll break it down step by step, and you'll see how manageable it really is. Our mission is to simplify a trigonometric expression without using a calculator. That's right, we're going old school and relying on our knowledge of trigonometric identities and relationships.
2. 1 Simplifying $\cos (-\alpha) \times \sin (90^{\circ}-\alpha) \times(1+\tan ^2 \alpha)$
Trigonometric expressions can often appear complex, but with the right approach and a solid understanding of trigonometric identities, simplifying them becomes a breeze. Let's embark on this journey of simplifying the expression $\cos(-\alpha) \times \sin(90^{\circ}-\alpha) \times (1 + \tan^2 \alpha)$. Our goal is to transform this expression into its simplest form without relying on a calculator. We'll achieve this by strategically applying fundamental trigonometric identities and properties. Remember, the key to success in trigonometry lies in recognizing patterns and knowing which identities to use when. So, let's roll up our sleeves and dive in!
First, let's start by recalling some key trigonometric identities that will be crucial in our simplification process. One of the first identities we'll use is the even-odd identity for cosine, which states that $\cos(-\alpha) = \cos(\alpha)$. This identity tells us that the cosine function is an even function, meaning it's symmetrical about the y-axis. Next, we'll utilize the co-function identity for sine, which states that $\sin(90^{\circ} - \alpha) = \cos(\alpha)$. This identity arises from the complementary angle relationship in a right-angled triangle. Finally, we'll employ the Pythagorean identity, which states that $1 + \tan^2(\alpha) = \sec^2(\alpha)$. This identity is a cornerstone of trigonometric simplification and is derived from the fundamental Pythagorean theorem. With these identities in our toolkit, we're well-equipped to simplify the given expression. So, let's put these identities into action and watch the expression transform before our eyes!
Now that we have our identities ready, let's apply them to the expression step by step. Starting with the first term, $\cos(-\alpha)$, we can directly apply the even-odd identity to simplify it to $\cos(\alpha)$. This simplifies our expression to $\cos(\alpha) \times \sin(90^{\circ} - \alpha) \times (1 + \tan^2 \alpha)$. Next, we focus on the second term, $\sin(90^{\circ} - \alpha)$. Using the co-function identity, we can replace this term with $\cos(\alpha)$. Our expression now looks like this: $\cos(\alpha) \times \cos(\alpha) \times (1 + \tan^2 \alpha)$. Notice how the expression is already becoming simpler! Finally, we address the last term, $1 + \tan^2(\alpha)$. Here, we use the Pythagorean identity to replace this term with $\sec^2(\alpha)$. Our expression is now: $\cos(\alpha) \times \cos(\alpha) \times \sec^2(\alpha)$. We're in the home stretch now! Remember, the key to simplifying trigonometric expressions is to keep applying identities until you reach the simplest possible form. So, let's see what we can do with this expression.
We've made significant progress in simplifying the expression, but we're not quite done yet. Let's take a closer look at our current expression: $\cos(\alpha) \times \cos(\alpha) \times \sec^2(\alpha)$. We can rewrite this as $\cos^2(\alpha) \times \sec^2(\alpha)$. Now, let's recall the relationship between cosine and secant. The secant function is the reciprocal of the cosine function, which means that $\sec(\alpha) = \frac1}{\cos(\alpha)}$. Therefore, $\sec^2(\alpha) = \frac{1}{\cos^2(\alpha)}$. Substituting this into our expression, we get\cos^2(\alpha)}$. Now, it's clear that we have a common factor in the numerator and the denominator, which we can cancel out. This leaves us with-\alpha) \times (1 + \tan^2 \alpha)$ to its simplest form, which is 1. This process highlights the power of trigonometric identities in simplifying complex expressions. By strategically applying these identities, we can transform seemingly complicated expressions into elegant and concise forms. So, let's celebrate our success and move on to the next challenge!
Therefore,
2. 2 Given: $\frac{1}{(\cos \theta+\sin \theta)(\cos \theta-\sin \theta)}=\frac{k}{1-2 \sin ^2 \theta}$,Determine the value of $k$.
Determining the value of k in a trigonometric equation involves manipulating the equation using trigonometric identities and algebraic techniques until we isolate $k$. The given equation is $\frac{1}{(\cos \theta + \sin \theta)(\cos \theta - \sin \theta)} = \frac{k}{1 - 2\sin^2 \theta}$. Our goal is to find the value of $k$ that makes this equation true for all values of $\theta$ (where the equation is defined). To achieve this, we will simplify both sides of the equation and then compare them. The left-hand side of the equation looks like a difference of squares in the denominator, which is a classic pattern to exploit. The right-hand side involves $1 - 2\sin^2 \theta$, which might ring a bell as a double-angle identity for cosine. With these clues in mind, let's dive into the simplification process and see how we can unravel the value of $k$.
Let's begin by focusing on the left-hand side of the equation: $\frac{1}{(\cos \theta + \sin \theta)(\cos \theta - \sin \theta)}$. Notice that the denominator is in the form of $(a + b)(a - b)$, which we know is equal to $a^2 - b^2$. Applying this algebraic identity, we can rewrite the denominator as $\cos^2 \theta - \sin^2 \theta$. So, the left-hand side now becomes $\frac{1}{\cos^2 \theta - \sin^2 \theta}$. This is a significant simplification, but we're not quite there yet. Now, let's turn our attention to the right-hand side of the equation and see if we can relate it to our simplified left-hand side. The right-hand side is given as $\frac{k}{1 - 2\sin^2 \theta}$. The denominator here, $1 - 2\sin^2 \theta$, looks suspiciously like a double-angle identity for cosine. In fact, we know that $\cos(2\theta) = 1 - 2\sin^2 \theta$. This connection between the two sides of the equation is a crucial breakthrough. It suggests that we're on the right track and that we can likely find the value of $k$ by equating the two sides.
Now that we've simplified both sides of the equation and recognized the double-angle identity, let's bring it all together. We have the left-hand side as $\frac1}{\cos^2 \theta - \sin^2 \theta}$ and the right-hand side as $\frac{k}{1 - 2\sin^2 \theta}$. We also know that $1 - 2\sin^2 \theta = \cos(2\theta)$. But wait, there's more! The expression $\cos^2 \theta - \sin^2 \theta$ is also a double-angle identity for cosine! Specifically, $\cos(2\theta) = \cos^2 \theta - \sin^2 \theta$. This is a fantastic realization because it means we can rewrite the left-hand side as $\frac{1}{\cos(2\theta)}$. Now, our equation looks like this{\cos(2\theta)} = \frac{k}{\cos(2\theta)}$. This is a much simpler equation to work with. To find the value of $k$, we can simply multiply both sides of the equation by $\cos(2\theta)$. This gives us: $1 = k$. And there you have it! We've successfully determined the value of $k$ to be 1. This solution highlights the importance of recognizing and applying trigonometric identities to simplify equations. By using the double-angle identity for cosine, we were able to transform a complex equation into a simple one, allowing us to easily solve for $k$. So, let's celebrate our achievement and move on to the next challenge!
In this exploration of trigonometric simplification, we've seen how a combination of key trigonometric identities and algebraic manipulation can transform complex expressions into simpler forms. We successfully simplified $\cos(-\alpha) \times \sin(90^{\circ}-\alpha) \times (1 + \tan^2 \alpha)$ to 1 and determined the value of $k$ in the equation $\frac{1}{(\cos \theta+\sin \theta)(\cos \theta-\sin \theta)} = \frac{k}{1-2 \sin ^2 \theta}$ to be 1. These examples showcase the power of understanding and applying trigonometric identities. So, keep practicing, keep exploring, and you'll become a master of trigonometric simplification in no time! Remember, the world of trigonometry is vast and fascinating, and there's always more to discover. Keep up the great work, guys!