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MTH 1120, College Trigonometry, Quiz 1 Name: Instructions: Work the following problems; give your reasoning where appropriate. Do not give decimal approximations. 1. If t is a real number, √ 1 1 =√ . 15, find sin t. Solution: sin t = csc t 15 √ 5 1 3 (b) and cot t = − , find tan t. Solution: tan t = = −√ . 3 cot t 5 (a) and csc t = 2. Find the values of all of the trigonometric functions for the real number t given that tan t = 15 − and P (t), the image of t under the action of the wrapping function, is in quadrant II. 8 Solution: If P (t) is in quadrant II and tan t = −15/8, then P (t) must have the form (x, y), where x2 +y 2 = 1 and y/x = −15/8. From the latter of these equations, we see that y = −15x/8. Replacing y with −15x/8 in the equation for the unit circle gives 15 1=x +y =x + − x 8 2 2 2 2 225 289 2 = 1+ x2 = x . 64 64 Hence x = ±8/17. But P (t) is in the second quadrant, so x = −8/17. From y = −15x/8, we 8 15 17 17 8 have y = 15/17. It follows that cos t = − ; sin t = ; sec t = − ; csc t = ; cot t = − . 17 17 8 15 15 3. You are given that P (t), the image of the real number t under the action of the wrapping function, lies in quadrant II, and that the point (−5, 12) lies on the half-line that extends from the origin through P (t). Find the values of the six trigonometric functions of t. Solution: We know that P (t) lies on the unit circle and on the line through (0, 0) and (−5, 12), whose equation is y = −12x/5. Thus, the coordinates of P (t) have the form (x, y), where x2 + y 2 = 1 and y = −12x/5. Substituting the latter into the former, we find that 12 1=x +y =x + − x 5 2 2 2 2 144 169 2 = 1+ x2 = x , 25 25 so that x = ±5/13. But P (t) lies in the second quadrant, so x = −5/13. Then y = −12x/5 = 5 12 12 13 13 12/13, and it follows that cos θ = − ; sin θ = ; tan θ = − ; sec θ = − ; csc θ = ; 13 13 5 5 12 5 cot θ = − . 12