Trigonometric Values: Solving For Exact Values

by Andrew McMorgan 47 views

Hey Plastik Magazine readers! Today, we're diving deep into the fascinating world of trigonometry. We're going to tackle a problem that involves finding the exact values of trigonometric functions, given some initial conditions. Don't worry if it sounds intimidating; we'll break it down step by step. Our specific task is to determine the exact values when we know sin⁡α=941{\sin \alpha = \frac{9}{41}} and 0<α<π2{0 < \alpha < \frac{\pi}{2}}, as well as cos⁡β=68585{\cos \beta = \frac{6\sqrt{85}}{85}} and −π2<β<0{-\frac{\pi}{2} < \beta < 0}. This involves using trigonometric identities and understanding the properties of trigonometric functions in different quadrants. So, let’s get started and unlock the secrets behind this trigonometric challenge!

Understanding the Problem

Before we jump into calculations, let's make sure we understand what the problem is asking. In essence, we're given the sine of an angle α{\alpha} and the cosine of an angle β{\beta}, along with the quadrants in which these angles lie. Knowing this, we need to find the exact values of other trigonometric functions for these angles, such as cosine of α{\alpha}, sine of β{\beta}, tangent of both angles, and possibly even more complex expressions involving these angles. This kind of problem is a staple in trigonometry and is super important for anyone studying math, physics, or engineering. The key here is that each trigonometric function is positive or negative in certain quadrants, and using this information combined with trigonometric identities, we can solve the problem systematically. So, keep your thinking caps on, and let’s dive into the nitty-gritty details!

Breaking Down the Given Information

Let’s start by dissecting the information we've been given. First off, we know that sin⁡α=941{\sin \alpha = \frac{9}{41}}. This tells us the ratio of the opposite side to the hypotenuse in a right-angled triangle where α{\alpha} is one of the acute angles. We're also given that 0<α<π2{0 < \alpha < \frac{\pi}{2}}, which means α{\alpha} lies in the first quadrant. In the first quadrant, all trigonometric functions (sine, cosine, tangent, etc.) are positive. This is super handy because it means we don't have to worry about negative signs when finding other values for α{\alpha}. Next, we have cos⁡β=68585{\cos \beta = \frac{6\sqrt{85}}{85}}. This gives us the ratio of the adjacent side to the hypotenuse for angle β{\beta}. The condition −π2<β<0{-\frac{\pi}{2} < \beta < 0} places β{\beta} in the fourth quadrant. In the fourth quadrant, cosine is positive, but sine is negative. This is a crucial piece of information because it will affect the sign of the sine value we calculate for β{\beta}. Understanding these details is like laying the foundation for a strong building; it makes the rest of the solution much easier to construct.

Finding cos⁡α{\cos \alpha} and sin⁡β{\sin \beta}

Now that we have a good grasp of the problem, let's get down to some actual calculations! Our first task is to find cos⁡α{\cos \alpha} and sin⁡β{\sin \beta}. To do this, we'll use the fundamental Pythagorean trigonometric identity, which states that sin⁡2θ+cos⁡2θ=1{\sin^2 \theta + \cos^2 \theta = 1} for any angle θ{\theta}. This identity is like the Swiss Army knife of trigonometry – super versatile and useful in many situations. For angle α{\alpha}, we already know sin⁡α=941{\sin \alpha = \frac{9}{41}}, so we can plug this into the identity and solve for cos⁡α{\cos \alpha}. Remember, since α{\alpha} is in the first quadrant, cos⁡α{\cos \alpha} will be positive. For angle β{\beta}, we know cos⁡β=68585{\cos \beta = \frac{6\sqrt{85}}{85}}, and we'll use the same identity to find sin⁡β{\sin \beta}. However, since β{\beta} is in the fourth quadrant, sin⁡β{\sin \beta} will be negative. Keeping track of these signs is key to getting the correct answers. So, let's roll up our sleeves and crunch some numbers!

Calculating cos⁡α{\cos \alpha}

Okay, let's calculate cos⁡α{\cos \alpha}. We start with the Pythagorean identity:

sin⁡2α+cos⁡2α=1\sin^2 \alpha + \cos^2 \alpha = 1

We know sin⁡α=941{\sin \alpha = \frac{9}{41}}, so we substitute that in:

(941)2+cos⁡2α=1\left(\frac{9}{41}\right)^2 + \cos^2 \alpha = 1

This simplifies to:

811681+cos⁡2α=1\frac{81}{1681} + \cos^2 \alpha = 1

Now, we subtract 811681{\frac{81}{1681}} from both sides to isolate cos⁡2α{\cos^2 \alpha}:

cos⁡2α=1−811681=1681−811681=16001681\cos^2 \alpha = 1 - \frac{81}{1681} = \frac{1681 - 81}{1681} = \frac{1600}{1681}

To find cos⁡α{\cos \alpha}, we take the square root of both sides:

cos⁡α=±16001681=±4041\cos \alpha = \pm \sqrt{\frac{1600}{1681}} = \pm \frac{40}{41}

Since α{\alpha} is in the first quadrant, cos⁡α{\cos \alpha} is positive, so we take the positive root:

cos⁡α=4041\cos \alpha = \frac{40}{41}

There we have it! We've successfully found cos⁡α{\cos \alpha}. This is a great example of how using trigonometric identities, combined with information about the quadrant, helps us nail down the exact value. Now, let's move on to finding sin⁡β{\sin \beta}.

Determining sin⁡β{\sin \beta}

Alright, next up, let's find sin⁡β{\sin \beta}. We'll use the same Pythagorean identity as before:

sin⁡2β+cos⁡2β=1\sin^2 \beta + \cos^2 \beta = 1

We know cos⁡β=68585{\cos \beta = \frac{6\sqrt{85}}{85}}, so we substitute that in:

sin⁡2β+(68585)2=1\sin^2 \beta + \left(\frac{6\sqrt{85}}{85}\right)^2 = 1

Squaring the cosine term, we get:

sin⁡2β+36⋅85852=1\sin^2 \beta + \frac{36 \cdot 85}{85^2} = 1

sin⁡2β+3685=1\sin^2 \beta + \frac{36}{85} = 1

Now, subtract 3685{\frac{36}{85}} from both sides:

sin⁡2β=1−3685=85−3685=4985\sin^2 \beta = 1 - \frac{36}{85} = \frac{85 - 36}{85} = \frac{49}{85}

Taking the square root of both sides gives us:

sin⁡β=±4985=±785\sin \beta = \pm \sqrt{\frac{49}{85}} = \pm \frac{7}{\sqrt{85}}

Now, we need to rationalize the denominator by multiplying the numerator and denominator by 85{\sqrt{85}}:

sin⁡β=±78585\sin \beta = \pm \frac{7\sqrt{85}}{85}

Since β{\beta} is in the fourth quadrant, sin⁡β{\sin \beta} is negative, so we choose the negative root:

sin⁡β=−78585\sin \beta = -\frac{7\sqrt{85}}{85}

Fantastic! We've found sin⁡β{\sin \beta}. Remember how important it was to consider the quadrant? If we had forgotten that sine is negative in the fourth quadrant, we would have ended up with the wrong sign. So, always keep those quadrant rules in mind!

Calculating Other Trigonometric Functions

Now that we've found cos⁡α{\cos \alpha} and sin⁡β{\sin \beta}, we can calculate other trigonometric functions for these angles. This is where things start to get really interesting! We can find the tangent, cotangent, secant, and cosecant for both α{\alpha} and β{\beta} using the values we've already found and some basic trigonometric definitions. For example, the tangent is the ratio of sine to cosine, cotangent is the reciprocal of tangent, secant is the reciprocal of cosine, and cosecant is the reciprocal of sine. These relationships are like the building blocks of trigonometry, and mastering them opens up a whole new world of problem-solving possibilities. So, let's put these definitions to work and find the remaining trigonometric values for our angles.

Finding tan⁡α{\tan \alpha}, cot⁡α{\cot \alpha}, sec⁡α{\sec \alpha}, and csc⁡α{\csc \alpha}

Let's start with angle α{\alpha}. We already know sin⁡α=941{\sin \alpha = \frac{9}{41}} and cos⁡α=4041{\cos \alpha = \frac{40}{41}}. Now we can find the other trigonometric functions:

  • Tangent (tan⁡α{\tan \alpha}):

    tan⁡α=sin⁡αcos⁡α=9414041=940\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{9}{41}}{\frac{40}{41}} = \frac{9}{40}

  • Cotangent (cot⁡α{\cot \alpha}):

    cot⁡α=1tan⁡α=1940=409\cot \alpha = \frac{1}{\tan \alpha} = \frac{1}{\frac{9}{40}} = \frac{40}{9}

  • Secant (sec⁡α{\sec \alpha}):

    sec⁡α=1cos⁡α=14041=4140\sec \alpha = \frac{1}{\cos \alpha} = \frac{1}{\frac{40}{41}} = \frac{41}{40}

  • Cosecant (csc⁡α{\csc \alpha}):

    csc⁡α=1sin⁡α=1941=419\csc \alpha = \frac{1}{\sin \alpha} = \frac{1}{\frac{9}{41}} = \frac{41}{9}

See how smoothly that went? Once we had sine and cosine, the rest just fell into place. This shows the power of knowing the fundamental trigonometric ratios and their relationships. Now, let's do the same for angle β{\beta}.

Determining tan⁡β{\tan \beta}, cot⁡β{\cot \beta}, sec⁡β{\sec \beta}, and csc⁡β{\csc \beta}

Now let's tackle angle β{\beta}. We know cos⁡β=68585{\cos \beta = \frac{6\sqrt{85}}{85}} and sin⁡β=−78585{\sin \beta = -\frac{7\sqrt{85}}{85}}. Let's find the remaining trig functions:

  • Tangent (tan⁡β{\tan \beta}):

    tan⁡β=sin⁡βcos⁡β=−7858568585=−76\tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{-\frac{7\sqrt{85}}{85}}{\frac{6\sqrt{85}}{85}} = -\frac{7}{6}

  • Cotangent (cot⁡β{\cot \beta}):

    cot⁡β=1tan⁡β=1−76=−67\cot \beta = \frac{1}{\tan \beta} = \frac{1}{-\frac{7}{6}} = -\frac{6}{7}

  • Secant (sec⁡β{\sec \beta}):

    sec⁡β=1cos⁡β=168585=85685\sec \beta = \frac{1}{\cos \beta} = \frac{1}{\frac{6\sqrt{85}}{85}} = \frac{85}{6\sqrt{85}}

    We'll rationalize the denominator by multiplying the numerator and denominator by 85{\sqrt{85}}:

    sec⁡β=85856⋅85=856\sec \beta = \frac{85\sqrt{85}}{6 \cdot 85} = \frac{\sqrt{85}}{6}

  • Cosecant (csc⁡β{\csc \beta}):

    csc⁡β=1sin⁡β=1−78585=−85785\csc \beta = \frac{1}{\sin \beta} = \frac{1}{-\frac{7\sqrt{85}}{85}} = -\frac{85}{7\sqrt{85}}

    Again, we rationalize the denominator:

    csc⁡β=−85857⋅85=−857\csc \beta = -\frac{85\sqrt{85}}{7 \cdot 85} = -\frac{\sqrt{85}}{7}

And there we have it! We've found all the primary trigonometric functions for both angles α{\alpha} and β{\beta}. This was a comprehensive exercise, and you guys have rocked it! Understanding how to find these values is a cornerstone of trigonometry. Now that we have these values, we can tackle even more complex problems, like finding the values of trigonometric functions of combined angles, which is something we might explore next time. Keep practicing, and you'll become trigonometry pros in no time!

Conclusion

Alright, guys, we've reached the end of our trigonometric journey for today! We started with sin⁡α=941{\sin \alpha = \frac{9}{41}} and cos⁡β=68585{\cos \beta = \frac{6\sqrt{85}}{85}}, along with their respective quadrants, and we successfully navigated our way to finding all the other trigonometric functions for both angles. We used the Pythagorean identity, understood the importance of quadrants, and applied the fundamental definitions of trigonometric functions. This whole process is a fantastic illustration of how trigonometry works and how different concepts fit together. The key takeaways here are: understanding the Pythagorean identity, knowing the signs of trigonometric functions in each quadrant, and being comfortable with the definitions of sine, cosine, tangent, and their reciprocals. With these tools in your mathematical toolkit, you'll be well-equipped to tackle a wide range of trigonometric problems. So, keep practicing, stay curious, and remember that every challenge is just an opportunity to learn something new. Until next time, keep those calculations sharp and your trigonometric minds engaged!