Solving Cos(π/4 - X) = (√2/2) Sin X Find Solutions

Hey everyone! Today, we're diving deep into a super interesting trigonometric problem that involves finding the solutions to the equation $\cos \left(\frac{\pi}{4}-x\right)=\frac{\sqrt{2}}{2} \sin x$ within the interval $[0, 2\pi]$. This problem isn't just about plugging in values; it's about understanding the relationships between trigonometric functions, applying identities, and ultimately, pinpointing the correct solutions. So, buckle up, and let's get started!

Unraveling the Trigonometric Equation

Our mission is clear: we need to find the values of x that satisfy the equation $\cos \left(\frac{\pi}{4}-x\right)=\frac{\sqrt{2}}{2} \sin x$ within the specified interval of $x \in [0, 2\pi]$. Now, this might seem a bit daunting at first, but don't worry, we'll break it down step by step. The key here is to use trigonometric identities to simplify the equation and make it easier to solve. We'll start by tackling the left side of the equation, which involves the cosine of a difference. This is where the cosine subtraction formula comes in handy. Remember, the cosine subtraction formula states that $\cos(A - B) = \cos A \cos B + \sin A \sin B$. Applying this formula to our equation, we get:

$\cos \left(\frac{\pi}{4}-x\right) = \cos \frac{\pi}{4} \cos x + \sin \frac{\pi}{4} \sin x$

Now, we know that $\cos \frac{\pi}{4}$ and $\sin \frac{\pi}{4}$ are both equal to $\frac{\sqrt{2}}{2}$. This is a crucial piece of information that simplifies our equation even further. Substituting these values, we have:

$\frac{\sqrt{2}}{2} \cos x + \frac{\sqrt{2}}{2} \sin x$

So, we've successfully expanded the left side of the equation. Now, let's bring the whole equation together. We started with $\cos \left(\frac{\pi}{4}-x\right)=\frac{\sqrt{2}}{2} \sin x$, and we've transformed the left side to $\frac{\sqrt{2}}{2} \cos x + \frac{\sqrt{2}}{2} \sin x$. This means our equation now looks like this:

$\frac{\sqrt{2}}{2} \cos x + \frac{\sqrt{2}}{2} \sin x = \frac{\sqrt{2}}{2} \sin x$

See how things are starting to simplify? We're making progress towards isolating the variable x and finding our solutions. Stay with me, guys; we're about to unravel this trigonometric mystery!

Simplifying and Solving the Equation

Alright, let's keep the momentum going! We've arrived at a simplified version of our original equation: $\frac{\sqrt{2}}{2} \cos x + \frac{\sqrt{2}}{2} \sin x = \frac{\sqrt{2}}{2} \sin x$. Now, the next step is to further simplify this equation to isolate the trigonometric functions and eventually solve for x. Notice that we have $\frac{\sqrt{2}}{2} \sin x$ on both sides of the equation. This is excellent news because we can subtract $\frac{\sqrt{2}}{2} \sin x$ from both sides, which will help us eliminate this term and simplify things considerably. Doing so, we get:

$\frac{\sqrt{2}}{2} \cos x = 0$

Wow, look at that! The equation has become much simpler. We're now dealing with just one trigonometric function, $\cos x$. To isolate $\cos x$, we can multiply both sides of the equation by $\frac{2}{\sqrt{2}}$. This might seem like a small step, but it's crucial for getting $\cos x$ by itself. Multiplying both sides, we have:

$\cos x = 0$

Now we're talking! We've successfully isolated $\cos x$. The equation $\cos x = 0$ is a fundamental trigonometric equation that we can solve by recalling the unit circle and the values of cosine at different angles. Remember, the cosine function corresponds to the x-coordinate on the unit circle. So, we're looking for the angles where the x-coordinate is zero. Within the interval $[0, 2\pi]$, there are two angles where this occurs. Can you guess what they are?

The angles are $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$. At these angles, the point on the unit circle is either (0, 1) or (0, -1), both of which have an x-coordinate of 0. So, we've found our solutions! But let's not stop here. It's always good to double-check our answers to make sure they're correct.

Verifying the Solutions

Okay, we've arrived at the potential solutions $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$. But before we declare victory, it's crucial to verify these solutions. This step is essential to ensure that our answers actually satisfy the original equation $\cos \left(\frac{\pi}{4}-x\right)=\frac{\sqrt{2}}{2} \sin x$. We'll plug each solution back into the original equation and see if both sides are equal.

Let's start with $x = \frac{\pi}{2}$. Substituting this value into the original equation, we get:

$\cos \left(\frac{\pi}{4}-\frac{\pi}{2}\right)=\frac{\sqrt{2}}{2} \sin \frac{\pi}{2}$

Now, we need to simplify both sides of this equation. First, let's simplify the angle inside the cosine function: $\frac{\pi}{4}-\frac{\pi}{2} = \frac{\pi}{4}-\frac{2\pi}{4} = -\frac{\pi}{4}$. So, the left side of the equation becomes $\cos \left(-\frac{\pi}{4}\right)$. Remember that cosine is an even function, meaning $\cos(-x) = \cos(x)$. Therefore, $\cos \left(-\frac{\pi}{4}\right) = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$.

Now, let's look at the right side of the equation: $\frac{\sqrt{2}}{2} \sin \frac{\pi}{2}$. We know that $\sin \frac{\pi}{2} = 1$, so the right side simplifies to $\frac{\sqrt{2}}{2} \cdot 1 = \frac{\sqrt{2}}{2}$.

Comparing both sides, we see that $\frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}$. This means that $x = \frac{\pi}{2}$ is indeed a valid solution. Great! We've verified one solution. Now, let's move on to the next one.

Next, we'll verify $x = \frac{3\pi}{2}$. Substituting this value into the original equation, we get:

$\cos \left(\frac{\pi}{4}-\frac{3\pi}{2}\right)=\frac{\sqrt{2}}{2} \sin \frac{3\pi}{2}$

Again, we need to simplify both sides. Let's start with the angle inside the cosine function: $\frac{\pi}{4}-\frac{3\pi}{2} = \frac{\pi}{4}-\frac{6\pi}{4} = -\frac{5\pi}{4}$. So, the left side becomes $\cos \left(-\frac{5\pi}{4}\right)$. Using the even function property of cosine, we have $\cos \left(-\frac{5\pi}{4}\right) = \cos \frac{5\pi}{4}$. Now, $\frac{5\pi}{4}$ is in the third quadrant, where both cosine and sine are negative. The reference angle for $\frac{5\pi}{4}$ is $\frac{\pi}{4}$, so $\cos \frac{5\pi}{4} = -\frac{\sqrt{2}}{2}$.

Now, let's look at the right side: $\frac{\sqrt{2}}{2} \sin \frac{3\pi}{2}$. We know that $\sin \frac{3\pi}{2} = -1$, so the right side simplifies to $\frac{\sqrt{2}}{2} \cdot (-1) = -\frac{\sqrt{2}}{2}$.

Comparing both sides, we see that $-\frac{\sqrt{2}}{2} = -\frac{\sqrt{2}}{2}$. This confirms that $x = \frac{3\pi}{2}$ is also a valid solution. Fantastic! We've verified both of our solutions.

Conclusion: The Solutions Unveiled

Alright, guys, we've reached the end of our trigonometric journey! We started with the equation $\cos \left(\frac{\pi}{4}-x\right)=\frac{\sqrt{2}}{2} \sin x$ and, through careful simplification, application of trigonometric identities, and diligent verification, we've successfully found the solutions within the interval $[0, 2\pi]$.

Our solutions are $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$. These are the values that make the original equation true. Remember, when tackling trigonometric equations, it's all about breaking down the problem into manageable steps, using the right identities, and always, always verifying your solutions.

So, to recap, we used the cosine subtraction formula, simplified the equation, isolated the cosine function, and then used our knowledge of the unit circle to find the angles where $\cos x = 0$. Finally, we verified our solutions by plugging them back into the original equation.

I hope this comprehensive guide has helped you understand how to solve this type of trigonometric problem. Keep practicing, and you'll become a trig whiz in no time! Until next time, keep exploring the fascinating world of mathematics!