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

To solve \( 1.1 \tan \theta=\frac{9}{40} \), we can rearrange it to find the sine and cosine values. First, \( \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{9}{40} \). Since \( \cos \theta < 0 \), we are looking at the second quadrant, where sine is positive and cosine is negative. Using the identity \( \tan^2 \theta + 1 = \sec^2 \theta \), we can first find \( \sin \theta \) and \( \cos \theta \): 1. Set \( \sin \theta = 9k \) and \( \cos \theta = -40k \), where \( k \) is a positive scaling factor. 2. Then \( \tan \theta = \frac{9}{-40} \), which is consistent with our quadrant determination. 3. Now using \( \sin^2 \theta + \cos^2 \theta = 1 \), we have \( (9k)^2 + (-40k)^2 = 1 \). 4. Solving this gives \( 81k^2 + 1600k^2 = 1 \) or \( 1681k^2 = 1 \). 5. Thus, \( k = \frac{1}{41} \). Now, substituting back, we find: \( \sin \theta = 9k = \frac{9}{41} \) and \( \cos \theta = -40k = -\frac{40}{41} \). Finally, for \( \sin \theta + \cos \theta \): \[ \sin \theta + \cos \theta = \frac{9}{41} - \frac{40}{41} = \frac{9 - 40}{41} = \frac{-31}{41}. \] So, the value of \( \sin \theta + \cos \theta \) is \( \frac{-31}{41} \). Now, as a side note, it’s interesting to think about where this all fits historically. The study of trigonometric functions dates back to ancient civilizations, with mathematicians like Hipparchus and later Islamic scholars developing tables and understanding of these ratios. They utilized geometry for astronomy and built a foundation for calculus and analysis! On the practical side, understanding these trigonometric ratios and their signs in different quadrants can significantly simplify physics problems, especially in mechanics where angles of projection or forces need resolution into components. So mastering this concept is not just for math class; it's essential across many real-world scenarios!