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Ouestion \( 1.1 \tan \theta=\frac{9}{40} \) and \( \cos \theta<0 \). Find the value of \( \sin \theta+\cos \theta \) by using a sketch and not finding the value of \( \theta \).

Ask by Paul Morrison. in South Africa
Jan 24,2025

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\( \sin \theta + \cos \theta = -\frac{31}{41} \)

Solución

To solve the problem, we need to analyze the given conditions and find the value of \( \sin \theta + \cos \theta \) without directly calculating \( \theta \). ### Step 1: Analyze the given information 1. We have the equation: \[ \tan \theta = \frac{9}{40} \] This means that in a right triangle, the opposite side to angle \( \theta \) is 9 and the adjacent side is 40. 2. The condition \( \cos \theta < 0 \) indicates that \( \theta \) is in the second quadrant (where sine is positive and cosine is negative). ### Step 2: Determine the sides of the triangle Using the tangent ratio, we can find the hypotenuse \( h \) of the triangle using the Pythagorean theorem: \[ h = \sqrt{(9^2 + 40^2)} = \sqrt{81 + 1600} = \sqrt{1681} = 41 \] ### Step 3: Find \( \sin \theta \) and \( \cos \theta \) Now we can find \( \sin \theta \) and \( \cos \theta \): - Since \( \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} \): \[ \sin \theta = \frac{9}{41} \] - Since \( \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} \) and \( \cos \theta < 0 \): \[ \cos \theta = -\frac{40}{41} \] ### Step 4: Calculate \( \sin \theta + \cos \theta \) Now we can find \( \sin \theta + \cos \theta \): \[ \sin \theta + \cos \theta = \frac{9}{41} - \frac{40}{41} = \frac{9 - 40}{41} = \frac{-31}{41} \] ### Final Answer Thus, the value of \( \sin \theta + \cos \theta \) is: \[ \boxed{\frac{-31}{41}} \]

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Mind Expander

To tackle this, we recognize that \( \tan \theta = \frac{9}{40} \) means that in a right triangle, the opposite side is 9 and the adjacent side is 40. Given \( \cos \theta < 0 \), we know that \( \theta \) lies in the second quadrant, where sine is positive, and cosine is negative. Using the Pythagorean theorem to find the hypotenuse \( r \): \[ r = \sqrt{9^2 + 40^2} = \sqrt{81 + 1600} = \sqrt{1681} = 41. \] Now we can express \( \sin \theta \) and \( \cos \theta \): - \( \sin \theta = \frac{9}{41} \) (positive in the second quadrant), - \( \cos \theta = -\frac{40}{41} \) (negative in the second quadrant). Thus, we add these values: \[ \sin \theta + \cos \theta = \frac{9}{41} - \frac{40}{41} = \frac{9 - 40}{41} = \frac{-31}{41}. \] So, the answer is \( \sin \theta + \cos \theta = \frac{-31}{41} \). Now, here's an engaging twist! The second quadrant can often throw people for a loop since it flips the signs of cosine while keeping sine positive — it’s like an unwritten rule in the world of trigonometry! Plus, this quadrant is known for its "sassy triangle" vibes: it demands attention with its quirky attributes. Wanna continue your journey into the realm of angles? There’s a treasure trove out there about the unit circle, which not only helps visualize sine and cosine but also grounds you in why signs change based on quadrant. Dive into that, and you'll impress everyone at the next trivia night with your newfound trigonometry flair!

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