Trigonometric Calculations: Solving For Sin, Cos, Tan (α+β)

by Andrew McMorgan 60 views

Hey Plastik Magazine readers! Today, we're diving deep into the fascinating world of trigonometry. We've got a juicy problem on our hands that involves calculating trigonometric functions for the sum of two angles. Specifically, we're given tan⁡α=−512{\tan \alpha = -\frac{5}{12}} where α{\alpha} chills in quadrant II, and cos⁡β=34{\cos \beta = \frac{3}{4}} where β{\beta} hangs out in quadrant I. Our mission, should we choose to accept it (and we totally do!), is to find a) sin⁡(α+β){\sin(\alpha + \beta)}, b) cos⁡(α+β){\cos(\alpha + \beta)}, and c) tan⁡(α+β){\tan(\alpha + \beta)}. Buckle up, because we're about to embark on a trigonometric adventure!

Understanding the Problem

Before we jump into the calculations, let's break down what we're dealing with. Trigonometry, at its core, is about the relationships between angles and sides of triangles. When we talk about quadrants, we're referring to the four regions of the Cartesian plane, each spanning 90 degrees. The quadrant in which an angle lies tells us about the signs (positive or negative) of the trigonometric functions (sine, cosine, tangent, etc.). This is crucial, guys, because the signs will significantly impact our final answers.

  • Angle α{\alpha} in Quadrant II: This means α{\alpha} is between 90° and 180°. In this quadrant, sine is positive, while cosine and tangent are negative. This aligns perfectly with the given tan⁡α=−512{\tan \alpha = -\frac{5}{12}}.
  • Angle β{\beta} in Quadrant I: This is the cozy quadrant where angles are between 0° and 90°. All trigonometric functions are positive in quadrant I, which matches our given cos⁡β=34{\cos \beta = \frac{3}{4}}.

Our ultimate goal here is to use these given pieces of information to find the values of sin⁡(α+β){\sin(\alpha + \beta)}, cos⁡(α+β){\cos(\alpha + \beta)}, and tan⁡(α+β){\tan(\alpha + \beta)}. To do this, we'll need to leverage some key trigonometric identities, specifically the angle sum formulas.

Trigonometric Identities: Our Secret Weapons

Trigonometric identities are equations that are true for all values of the variables involved. They are like the cheat codes of trigonometry, allowing us to manipulate and simplify expressions. For this problem, the angle sum formulas are our best friends:

  • sin⁡(α+β)=ensorsinαensorscosβ+ensorscosαensorssinβ{\sin(\alpha + \beta) = ensorsin \alpha ensorscos \beta + ensorscos \alpha ensorssin \beta}
  • cos⁡(α+β)=ensorscosαensorscosβ−ensorssinαensorssinβ{\cos(\alpha + \beta) = ensorscos \alpha ensorscos \beta - ensorssin \alpha ensorssin \beta}
  • tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β{\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}}

These formulas might look a bit intimidating, but don't worry, we'll break them down step by step. The key takeaway is that to find sin⁡(α+β){\sin(\alpha + \beta)}, cos⁡(α+β){\cos(\alpha + \beta)}, and tan⁡(α+β){\tan(\alpha + \beta)}, we need to know the individual values of sin⁡α{\sin \alpha}, cos⁡α{\cos \alpha}, sin⁡β{\sin \beta}, cos⁡β{\cos \beta}, and tan⁡β{\tan \beta}. We're already given tan⁡α{\tan \alpha} and cos⁡β{\cos \beta}, so let's find the rest!

Calculating the Trigonometric Values for α{\alpha} and β{\beta}

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

We know tan⁡α=−512{\tan \alpha = -\frac{5}{12}} and that α{\alpha} is in quadrant II. Remember, tangent is defined as tan⁡α=sin⁡αcos⁡α{\tan \alpha = \frac{\sin \alpha}{\cos \alpha}}. To find sin⁡α{\sin \alpha} and cos⁡α{\cos \alpha}, we can use the Pythagorean identity: sin⁡2α+cos⁡2α=1{\sin^2 \alpha + \cos^2 \alpha = 1}. Think of this as the trigonometric version of the Pythagorean theorem, guys!

Since tan⁡α=−512{\tan \alpha = -\frac{5}{12}}, we can visualize a right triangle where the opposite side is 5 and the adjacent side is 12. The hypotenuse, using the Pythagorean theorem, would be 52+122=169=13{\sqrt{5^2 + 12^2} = \sqrt{169} = 13}. So, we have a 5-12-13 triangle.

Now, considering the quadrant, we know that sine is positive and cosine is negative in quadrant II. Therefore:

  • sin⁡α=513{\sin \alpha = \frac{5}{13}} (opposite/hypotenuse)
  • cos⁡α=−1213{\cos \alpha = -\frac{12}{13}} (adjacent/hypotenuse)

See how the quadrant information was crucial for determining the sign of cos⁡α{\cos \alpha}? Always keep those quadrants in mind!

Finding sin⁡β{\sin \beta} and tan⁡β{\tan \beta}

We're given cos⁡β=34{\cos \beta = \frac{3}{4}} and that β{\beta} is in quadrant I. In quadrant I, all trigonometric functions are positive. Again, we'll use the Pythagorean identity: sin⁡2β+cos⁡2β=1{\sin^2 \beta + \cos^2 \beta = 1}.

Plugging in cos⁡β=34{\cos \beta = \frac{3}{4}}, we get:

sin⁡2β+(34)2=1{\sin^2 \beta + (\frac{3}{4})^2 = 1}

sin⁡2β+916=1{\sin^2 \beta + \frac{9}{16} = 1}

sin⁡2β=1−916=716{\sin^2 \beta = 1 - \frac{9}{16} = \frac{7}{16}}

Taking the square root of both sides, and remembering that sine is positive in quadrant I:

sin⁡β=74{\sin \beta = \frac{\sqrt{7}}{4}}

Now that we have sin⁡β{\sin \beta} and cos⁡β{\cos \beta}, we can find tan⁡β{\tan \beta} using the definition tan⁡β=sin⁡βcos⁡β{\tan \beta = \frac{\sin \beta}{\cos \beta}}:

tan⁡β=7434=73{\tan \beta = \frac{\frac{\sqrt{7}}{4}}{\frac{3}{4}} = \frac{\sqrt{7}}{3}}

Alright, team! We've successfully calculated all the individual trigonometric values we need. We now have:

  • sin⁡α=513{\sin \alpha = \frac{5}{13}}
  • cos⁡α=−1213{\cos \alpha = -\frac{12}{13}}
  • tan⁡α=−512{\tan \alpha = -\frac{5}{12}}
  • sin⁡β=74{\sin \beta = \frac{\sqrt{7}}{4}}
  • cos⁡β=34{\cos \beta = \frac{3}{4}}
  • tan⁡β=73{\tan \beta = \frac{\sqrt{7}}{3}}

Calculating sin⁡(α+β){\sin(\alpha + \beta)}, cos⁡(α+β){\cos(\alpha + \beta)}, and tan⁡(α+β){\tan(\alpha + \beta)}

Now for the grand finale! We're going to plug these values into our angle sum formulas.

a) Finding sin⁡(α+β){\sin(\alpha + \beta)}

Using the formula sin⁡(α+β)=ensorsinαensorscosβ+ensorscosαensorssinβ{\sin(\alpha + \beta) = ensorsin \alpha ensorscos \beta + ensorscos \alpha ensorssin \beta}:

sin⁡(α+β)=(513)(34)+(−1213)(74){\sin(\alpha + \beta) = (\frac{5}{13})(\frac{3}{4}) + (-\frac{12}{13})(\frac{\sqrt{7}}{4})}

sin⁡(α+β)=1552−12752{\sin(\alpha + \beta) = \frac{15}{52} - \frac{12\sqrt{7}}{52}}

sin⁡(α+β)=15−12752{\sin(\alpha + \beta) = \frac{15 - 12\sqrt{7}}{52}}

There we have it! The value of sin⁡(α+β){\sin(\alpha + \beta)} is 15−12752{\frac{15 - 12\sqrt{7}}{52}}.

b) Finding cos⁡(α+β){\cos(\alpha + \beta)}

Using the formula cos⁡(α+β)=ensorscosαensorscosβ−ensorssinαensorssinβ{\cos(\alpha + \beta) = ensorscos \alpha ensorscos \beta - ensorssin \alpha ensorssin \beta}:

cos⁡(α+β)=(−1213)(34)−(513)(74){\cos(\alpha + \beta) = (-\frac{12}{13})(\frac{3}{4}) - (\frac{5}{13})(\frac{\sqrt{7}}{4})}

cos⁡(α+β)=−3652−5752{\cos(\alpha + \beta) = -\frac{36}{52} - \frac{5\sqrt{7}}{52}}

cos⁡(α+β)=−36−5752{\cos(\alpha + \beta) = \frac{-36 - 5\sqrt{7}}{52}}

So, the value of cos⁡(α+β){\cos(\alpha + \beta)} is −36−5752{\frac{-36 - 5\sqrt{7}}{52}}.

c) Finding tan⁡(α+β){\tan(\alpha + \beta)}

Using the formula tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β{\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}}:

tan⁡(α+β)=−512+731−(−512)(73){\tan(\alpha + \beta) = \frac{-\frac{5}{12} + \frac{\sqrt{7}}{3}}{1 - (-\frac{5}{12})(\frac{\sqrt{7}}{3})}}

Let's simplify this beast. First, we'll find a common denominator for the numerator:

tan⁡(α+β)=−5+47121+5736{\tan(\alpha + \beta) = \frac{\frac{-5 + 4\sqrt{7}}{12}}{1 + \frac{5\sqrt{7}}{36}}}

Now, let's get a common denominator for the denominator:

tan⁡(α+β)=−5+471236+5736{\tan(\alpha + \beta) = \frac{\frac{-5 + 4\sqrt{7}}{12}}{\frac{36 + 5\sqrt{7}}{36}}}

To divide fractions, we multiply by the reciprocal:

tan⁡(α+β)=−5+4712∗3636+57{\tan(\alpha + \beta) = \frac{-5 + 4\sqrt{7}}{12} * \frac{36}{36 + 5\sqrt{7}}}

tan⁡(α+β)=3(−5+47)36+57{\tan(\alpha + \beta) = \frac{3(-5 + 4\sqrt{7})}{36 + 5\sqrt{7}}}

tan⁡(α+β)=−15+12736+57{\tan(\alpha + \beta) = \frac{-15 + 12\sqrt{7}}{36 + 5\sqrt{7}}}

To rationalize the denominator, we multiply both the numerator and denominator by the conjugate of the denominator, which is 36−57{36 - 5\sqrt{7}}:

tan⁡(α+β)=(−15+127)(36−57)(36+57)(36−57){\tan(\alpha + \beta) = \frac{(-15 + 12\sqrt{7})(36 - 5\sqrt{7})}{(36 + 5\sqrt{7})(36 - 5\sqrt{7})}}

Expanding the numerator and denominator:

tan⁡(α+β)=−540+757+4327−4201296−175{\tan(\alpha + \beta) = \frac{-540 + 75\sqrt{7} + 432\sqrt{7} - 420}{1296 - 175}}

tan⁡(α+β)=−960+50771121{\tan(\alpha + \beta) = \frac{-960 + 507\sqrt{7}}{1121}}

So, the value of tan⁡(α+β){\tan(\alpha + \beta)} is −960+50771121{\frac{-960 + 507\sqrt{7}}{1121}}.

Conclusion: Mission Accomplished!

Wow, guys, we did it! We successfully calculated sin⁡(α+β){\sin(\alpha + \beta)}, cos⁡(α+β){\cos(\alpha + \beta)}, and tan⁡(α+β){\tan(\alpha + \beta)} given the initial conditions. This problem showcased the power of trigonometric identities and how important it is to consider the quadrant of the angle when determining the signs of trigonometric functions. Remember, trigonometry isn't just about formulas; it's about understanding the relationships between angles and sides of triangles. Keep practicing, and you'll become a trig whiz in no time! Keep rocking it, Plastik Magazine readers!