Trigonometric graphs, identities and general triangles
ACT Topic 13 4:55 English narration · English + 中文 subtitles burned in
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Transcript
Amplitude is half the peak-to-trough distance.
振幅是波峰到波谷距离的一半。
In y equals A sine of B x plus D, it is the absolute value of A.
在 y = A sin(Bx) + D 中，振幅为 |A|。
Radian period is 2 pi divided by the absolute value of B.
弧度周期为 2π 除以 |B|。
For a non-right triangle, match known sides and included angle before selecting the cosine rule.
对于非直角三角形，在选择余弦定理之前，需先匹配已知边及其夹角。
C squared equals A squared plus B squared minus 2 A B cosine of C.
c² = a² + b² − 2ab cos C。
Each task gives its own data.
每个任务都有其自身的数据。
Do not copy the worked example’s numbers.
请勿照搬例题中的数值。
Pause here and attempt the task before revealing the reasoning.
请先暂停，完成这道原创练习再继续。
Use the full context on the preceding pages.
请利用前几页的完整背景信息。
The amplitude is 4, the period is 2 pi over 2, which is pi, and the range is from negative 1 to 7.
振幅是四，周期是二 π 除以二，即 π，值域是负一到七。
At x equals pi over 4, y equals negative 4 times sine of pi over 2, plus 3, which is negative 1.
当 x 等于四分之 π 时，y 等于负四乘 sin 二分之 π 再加三，等于负一。
The negative coefficient reflects the wave; it does not make the amplitude negative.
负系数让波形翻转，但不会让振幅变成负数。
Pause here and attempt the task before revealing the reasoning.
请先暂停，完成这道原创练习再继续。
Use the full context on the preceding pages.
请利用前几页的完整背景信息。
By the cosine rule, c squared equals a squared plus b squared minus 2 a b cos C, which is 6 squared plus 10 squared, minus 2 times 6 times 10 times cos 60 degrees, giving 76 square centimetres, so c equals 2 root 19 centimetres.
由余弦定理，c 的平方等于 a 平方加 b 平方减二 ab 乘 cos C，即六的平方加十的平方，减去二乘六乘十乘 cos 六十度，得到七十六平方厘米，所以 c 等于二倍根号十九厘米。
The area is a b sin C over 2, which is 6 times 10 times sin 60 degrees, over 2, giving 15 root 3 square centimetres.
面积等于 ab 乘 sin C 再除以二，即六乘十乘 sin 六十度除以二，得到十五倍根号三平方厘米。
Pause here and attempt the task before revealing the reasoning.
请先暂停，完成这道原创练习再继续。
Use the full context on the preceding pages.
请利用前几页的完整背景信息。
The angles are 30 and 150 degrees.
两个角分别是三十度和一百五十度。
Their cosines are root 3 over 2 and negative root 3 over 2; their tangents are one over root 3 and negative one over root 3.
它们的余弦是二分之根号三和负二分之根号三；正切是根号三分之一和负根号三分之一。
Sine is positive in quadrants one and two, so a single acute answer would be incomplete.
正弦在第一和第二象限都为正，所以只给一个锐角答案是不完整的。
Pause here and attempt the task before revealing the reasoning.
请先暂停，完成这道原创练习再继续。
Use the full context on the preceding pages.
请结合前几页的完整背景信息作答。
c is the square root of 9 squared plus 12 squared, which is 15 centimetres.
c 等于九的平方加十二的平方再开方，即十五厘米。
So sine theta is 9 over 15, or three fifths, and cos theta is 12 over 15, or four fifths; their squares sum to 1.
所以 sin θ 等于十五分之九，即五分之三；cos θ 等于十五分之十二，即五分之四；两者的平方和等于一。
In the second triangle, a known opposite side and angle pair permits b over sin 45 degrees equals 8 centimetres over sin 30 degrees.
在第二个三角形中，已知一组对边和对角，就可以写出 b 除以 sin 四十五度等于八厘米除以 sin 三十度。
Hence b equals 8 times root 2 over 2, divided by one half, which is 8 root 2 centimetres.
因此 b 等于八乘二分之根号二再除以二分之一，即八倍根号二厘米。
This triangle is not assumed right-angled; its third angle is 105°.
这个三角形并没有假定是直角三角形；它的第三个角是一百零五度。
Return to the task and explain why the method fits.
回到该任务，并解释为何此方法适用。
Original practice does not predict an official score.
原始练习无法预测正式成绩。
Complete required live or recorded evidence separately.
请单独完成所需的现场或录制证据。
Pause and test yourself. Cover the worked answer, rebuild the answer to the last example in your own words, and name one condition that would change it before you check.
暂停一下，自我检测：遮住示范答案，用自己的话重做最后一个例子的答案，并在核对之前说出一个会改变结论的条件。