Finding Cosine In Quadrant II

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In this article, we'll walk through the process of finding the value of $\cos(\theta)$ given that $\tan(\theta) = -\sqrt{\frac{19}{17}}$ and $ heta$ is an angle in quadrant II. This is a classic trigonometry problem that combines understanding trigonometric identities, quadrant rules, and algebraic manipulation. Let's dive in!

Understanding the Problem

Before we start solving, let's break down the problem. We are given that $\tan(\theta) = -\sqrt{\frac{19}{17}}$. The negative sign here is crucial because it tells us something about the possible quadrants in which the angle $\theta$ lies. Recall the acronym ASTC (All Students Take Calculus) or CAST, which helps us remember which trigonometric functions are positive in each quadrant:

  • Quadrant I (All): All trigonometric functions (sine, cosine, tangent) are positive.
  • Quadrant II (Sine): Only sine (and its reciprocal, cosecant) is positive.
  • Quadrant III (Tangent): Only tangent (and its reciprocal, cotangent) is positive.
  • Quadrant IV (Cosine): Only cosine (and its reciprocal, secant) is positive.

Since we are given that $\tan(\theta)$ is negative and that $\theta$ is in quadrant II, this information is consistent. In quadrant II, only sine is positive, while cosine and tangent are negative.

Our goal is to find $\cos(\theta)$. To do this, we'll use trigonometric identities, specifically the Pythagorean identity, and the relationship between tangent, sine, and cosine.

Utilizing Trigonometric Identities

The key identity we'll use is the Pythagorean identity:

sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1

We also know that tangent is the ratio of sine to cosine:

tan⁡(θ)=sin⁡(θ)cos⁡(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}

Our strategy will be to use the given value of $\tan(\theta)$ to find a relationship between $\sin(\theta)$ and $\cos(\theta)$, and then use the Pythagorean identity to solve for $\cos(\theta)$. Since we know we're in quadrant II, we'll also need to consider the sign of $\cos(\theta)$, which will be negative.

Step 1: Relate Tangent to Sine and Cosine

We have $\tan(\theta) = -\sqrt{\frac{19}{17}}$. This means:

sin⁡(θ)cos⁡(θ)=−1917\frac{\sin(\theta)}{\cos(\theta)} = -\sqrt{\frac{19}{17}}

To make things easier, let's square both sides of the equation:

(sin⁡(θ)cos⁡(θ))2=(−1917)2\left(\frac{\sin(\theta)}{\cos(\theta)}\right)^2 = \left(-\sqrt{\frac{19}{17}}\right)^2

This simplifies to:

sin⁡2(θ)cos⁡2(θ)=1917\frac{\sin^2(\theta)}{\cos^2(\theta)} = \frac{19}{17}

This equation gives us a relationship between $\sin^2(\theta)$ and $\cos^2(\theta)$. Now, we'll use the Pythagorean identity to eliminate $\sin^2(\theta)$.

Step 2: Use the Pythagorean Identity

From the Pythagorean identity, we have:

sin⁡2(θ)=1−cos⁡2(θ)\sin^2(\theta) = 1 - \cos^2(\theta)

Substitute this into our equation from Step 1:

1−cos⁡2(θ)cos⁡2(θ)=1917\frac{1 - \cos^2(\theta)}{\cos^2(\theta)} = \frac{19}{17}

Now we have an equation solely in terms of $\cos^2(\theta)$. Let's solve for it.

Step 3: Solve for Cosine Squared

To solve for $\cos^2(\theta)$, we first multiply both sides by $\cos^2(\theta)$:

1−cos⁡2(θ)=1917cos⁡2(θ)1 - \cos^2(\theta) = \frac{19}{17} \cos^2(\theta)

Next, add $\cos^2(\theta)$ to both sides:

1=1917cos⁡2(θ)+cos⁡2(θ)1 = \frac{19}{17} \cos^2(\theta) + \cos^2(\theta)

Combine the terms on the right side:

1=(1917+1)cos⁡2(θ)1 = \left(\frac{19}{17} + 1\right) \cos^2(\theta)

1=3617cos⁡2(θ)1 = \frac{36}{17} \cos^2(\theta)

Now, multiply both sides by $\frac{17}{36}$ to isolate $\cos^2(\theta)$:

cos⁡2(θ)=1736\cos^2(\theta) = \frac{17}{36}

Step 4: Solve for Cosine

To find $\cos(\theta)$, we take the square root of both sides:

cos⁡(θ)=±1736\cos(\theta) = \pm \sqrt{\frac{17}{36}}

cos⁡(θ)=±176\cos(\theta) = \pm \frac{\sqrt{17}}{6}

Now, we need to consider the sign. Since $\theta$ is in quadrant II, cosine is negative. Therefore:

cos⁡(θ)=−176\cos(\theta) = -\frac{\sqrt{17}}{6}

Verify the Answer

To ensure our answer is correct, let's verify it by plugging it back into our original equations and conditions.

We found that $\cos(\theta) = -\frac{\sqrt{17}}{6}$. We also know that $\tan(\theta) = -\sqrt{\frac{19}{17}}$. We can use these values to find $\sin(\theta)$ and check if everything is consistent.

First, let's find $\sin(\theta)$ using the tangent equation:

tan⁡(θ)=sin⁡(θ)cos⁡(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}

−1917=sin⁡(θ)−176-\sqrt{\frac{19}{17}} = \frac{\sin(\theta)}{-\frac{\sqrt{17}}{6}}

Multiply both sides by $-\frac{\sqrt{17}}{6}$:

sin⁡(θ)=−1917⋅(−176)\sin(\theta) = -\sqrt{\frac{19}{17}} \cdot \left(-\frac{\sqrt{17}}{6}\right)

sin⁡(θ)=1917⋅176\sin(\theta) = \sqrt{\frac{19}{17}} \cdot \frac{\sqrt{17}}{6}

sin⁡(θ)=196\sin(\theta) = \frac{\sqrt{19}}{6}

Now, let's check if our values satisfy the Pythagorean identity:

sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1

(196)2+(−176)2=1\left(\frac{\sqrt{19}}{6}\right)^2 + \left(-\frac{\sqrt{17}}{6}\right)^2 = 1

1936+1736=1\frac{19}{36} + \frac{17}{36} = 1

3636=1\frac{36}{36} = 1

1=11 = 1

The identity holds true. Also, since $\sin(\theta) = \frac{\sqrt{19}}{6}$ is positive, this is consistent with $\theta$ being in quadrant II.

Conclusion

By using the given information about $\tan(\theta)$ and the quadrant of $\theta$, along with trigonometric identities, we successfully found the value of $\cos(\theta)$. Remember the importance of the ASTC/CAST rule for determining the signs of trigonometric functions in different quadrants, and the power of the Pythagorean identity in solving trigonometric problems. Understanding these concepts will help you tackle a wide range of trigonometry questions.

In summary, the correct value for $\cos(\theta)$ is:

cos⁡(θ)=−176\cos(\theta) = -\frac{\sqrt{17}}{6}

This step-by-step approach can be applied to similar problems involving different trigonometric functions and quadrants. Practice and a solid understanding of the fundamentals are key to mastering trigonometry.

Keywords

  • Trigonometry
  • Cosine
  • Tangent
  • Pythagorean Identity
  • Quadrants
  • ASTC Rule
  • CAST Rule
  • Trigonometric Functions
  • Sine
  • Angle

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