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Solve .
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Question text Each student below analyzes a proposed trigonometric expression and explains their reasoning.Match each student’s reasoning to the explanation that best fits their thinking. Student 1:I was asked to evaluate .I tested it using and , and the two sides were not equal. I tried other values too and got the same result. I believe this expression is never true.Answer 1 Question 23[select: , The student was correct; the expression is never true. Multiple test cases using different values always yield unequal results., The student analysis was incorrect; the expression is always true because it is a valid trigonometric identity., The student analysis was incorrect; the expression is sometimes true. It holds for some distinct values of x and y, but is unequal for other values.] Student 2:I was asked to verify .I recognized the numerator as based on the Pythagorean identity, so the left side of the expression becomes .I now realize this means the expression is never true.Answer 2 Question 23[select: , The student's analysis was incorrect; the expression is always true because the numerator is equivalent to sin^2(x), not -sin^2(x)., The student's analysis was incorrect; the expression is sometimes true because it holds for specific values of x, but fails for most other values., The student was correct; the expression is never true since the left side simplifies to -1, which is never equal to the right side.] Student 3:I was asked to evaluate .I tested it using , and both sides came out equal. I concluded the expression is always true.Answer 3 Question 23[select: , The student was correct; the expression is always true for all values of x ., The student analysis was incorrect; the expression is sometimes true; when different values for x are tested, the two sides are not equal., The student analysis was incorrect; the expression is never true for any value of x.]
Question text The students below each construct a sinusoidal function based on a shared scenario involving water level fluctuations.Match each student’s reasoning to the explanation that best fits their thinking. Scenario:The depth of water , in metres, at a canal dock fluctuates due to the system of canal locks lowering and raising the water for container ships. The water varies sinusoidally from a minimum of 1.5 m to a maximum of 4.5 m. The depth reaches its minimum at 5:00 AM, and one full cycle of the water level is completed every 4 hours. The horizontal axis represents time , in hours, where corresponds to midnight. Student 1:I was asked to find the water depth at 3 AM.The equation I created was Since the minimum occurs at 5:00 AM, I shifted the function by 5.Using this function, I found the depth at 3 AM was 1.5 m.Answer 1 Question 22[select: , All parameters are correct. Therefore, the initial evaluation is correct, and the correct depth at 3 AM is 1.5 m. , All parameters are correct except the phase shift: there should be a phase shift of -5. Therefore, the initial evaluation is incorrect; using the new equation, the correct depth at 3 AM is 4.5 m. , All parameters are correct except the k-value: the k-value should be pi/4. The initial evaluation is still correct; the depth at 3 AM is 1.5 m.] Student 2:I was asked to find the water depth at 6 AM.The equation I created was I chose sine because the water starts rising after its lowest point.Using my equation, I found the depth at 6 AM was 3 m.Answer 2 Question 22[select: , All parameters are correct except for the amplitude, which should be negative. The initial evaluation is still correct; the depth at 6 AM is 3 m., All parameters are except the phase shift: there should be a phase shift of -5. Therefore, the initial evaluation is incorrect; using the new equation, the correct depth at 6 AM is 4.1 m., All parameters are correct except the k-value; the k-value should be pi/2. The initial evaluation is still correct; the depth at 6 AM is 3 m. ] Student 3:I was asked to find the water depth at 1 PM.The equation I created was Since the minimum occurs at 5:00 AM, I used a cosine function with a shift to match the point.I found the depth at 1 PM to be 4.5 m.Answer 3 Question 22[select: , All parameters are correct except the k-value; the correct k-value is pi/4. The initial evaluation is still correct; the depth at 1 PM is 4.5 m., All parameters are correct except for the phase shift; there should be a phase shift of -3. Therefore, the initial evaluation is incorrect; using the new equation, the correct depth at 1 PM is 1.5 m.. , All parameters are correct and the student’s evaluation of the depth was correct; the depth at 1 PM is 4.5 m.]
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