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Задача по математике, 9 класс: sine and cosine. Пошаговые подсказки, полное решение и ответ. Темы: Sine and cosine in a right triangle and on the unit semicircle. Длительность — 20 мин.
Условие задачи
Подсказки для объяснения
Открывайте подсказки постепенно — сначала дайте ребенку подумать самостоятельно.
Подсказка 1
In a right triangle, for angle A, identify the opposite and adjacent legs. What are the sine and cosine values? Does the answer change if you rotate the sheet?
Подсказка 2
How do the coordinates of a point change under reflection across the y-axis? Which angle on the unit circle corresponds to the new point?
Подсказка 3
What coordinates do the points on the upper semicircle have for angles 0°, 90°, and 180°? What does this mean for cos and sin?
Подсказка 4
For two equal sides and different included angles, compare the third side using the law of cosines. Which angle gives the largest side?
Подсказка 5
Write the law of cosines for sides 5 and 8 and angle 60°. What is the length of the third side? Does the triangle inequality hold?
Подсказка 6
Apply the law of cosines for sides 3 and 5 and angle 120°. What sign does cos 120° have? What is the length of the third side?
Подсказка 7
In the law of cosines for angle 120°, there must be a minus sign before the product of the sides. Which sign should replace the mistake?
Подсказка 8
How do sine and cosine change when moving from 40° to 140°? What signs will the values have in quadrant II?
Подсказка 9
Let x be the number of points for one like and y be the number of points for one repost. From the conditions 3x+2y=17 and x+4y=19, solve the system in two ways: substitution and elimination. Then check the result by substitution.
Пошаговое решение
1) In right triangle ABC with angle C=90°, for angle A the opposite leg is BC=4 cm, the adjacent leg is AC=3 cm, and the hypotenuse is AB=5 cm. Therefore sin A=BC/AB=4/5 and cos A=AC/AB=3/5. Rotating the sheet does not change the values of sine and cosine. 2) The point for 60° on the unit semicircle has coordinates (1/2; √3/2). Reflection across the y-axis changes the sign of the x-coordinate: we get (−1/2; √3/2). This is the point for angle 120°. Therefore sin 120°=√3/2 and cos 120°=−1/2. 3) On the unit semicircle, the point for 0° has coordinates (1;0), for 90° it is (0;1), and for 180° it is (−1;0). Since on the unit circle cos corresponds to the x-coordinate and sin to the y-coordinate, the pairs (cos;sin) are (1;0), (0;1), and (−1;0). 4) For two sides of 5 cm and angles 60°, 90°, and 120°, the third side increases as the included angle increases. So the order is: 60° — 5 cm, 90° — 5√2 cm, 120° — 5√3 cm. 5) By the law of cosines: a²=5²+8²−2·5·8·cos60°=25+64−80·(1/2)=49. Hence a=7 cm. Triangle inequality check: |8−5|<7<8+5, i.e. 3<7<13. 6) By the law of cosines: a²=3²+5²−2·3·5·cos120°=9+25−30·(−1/2)=49. So a=7 cm. Check: |5−3|<7<5+3, i.e. 2<7<8. 7) The sign is wrong: for 120° the cosine is negative, so the formula should not subtract 24; it should add 24. Correctly: a²=16+36−48·(−1/2)=76, so a=2√19 cm. 8) Since 140°=180°−40°, we have cos140°=−cos40°=−k and sin140°=sin40°=m. On the diagram this corresponds to moving into quadrant II: the x-coordinate changes sign, while the y-coordinate stays positive. 9) Let x be the points for one like and y be the points for one repost. We get the system: 3x+2y=17 and x+4y=19. Substitution method: from x+4y=19, x=19−4y. Substitute into the first equation: 3(19−4y)+2y=17, 57−12y+2y=17, 57−10y=17, −10y=−40, so y=4. Then x=19−4·4=3. Elimination method: multiply the second equation by 3: 3x+12y=57. Subtract the first equation 3x+2y=17: 10y=40, so y=4. Then x=3. Check by substitution: 3·3+2·4=9+8=17 and 1·3+4·4=3+16=19. So one like gives 3 points and one repost gives 4 points.
