Students learn sin(180° − θ) = sin θ and then apply the shape of it to the cosine. The fix is not a second rule to memorise; it is to stop memorising and read the reflection off the circle.
Set the slider to 60°, press the mirror button, and say nothing. Ask what moved. The answer is in front of them: the horizontal leg jumped to the other side of the y-axis and the height did not move at all. 0.866 both times, 0.5 becoming −0.5.
Then sweep the slider slowly through 90° and watch the cosine leg shrink to nothing, reverse, and grow the other way, while the sine leg is at its longest and barely changing. That is the whole second quadrant in one motion, and it is worth more than the table.
At 90° and 270° the tangent readout says undefined rather than a huge number. Stop there and ask why: the dashed line is vertical, and a vertical line has no gradient. Students who have only ever seen "tan 90 is infinity" on a calculator benefit from seeing a figure that refuses to print a number.
| Question | Answer |
|---|---|
| 1. sin 120° | 0.866, the same as sin 60°. Exactly √3/2. |
| 2. cos 120° | −0.5, which is −cos 60°. |
| 3. tan 210° | 0.5774, and POSITIVE. Exactly 1/√3. |
| 4. The other B | 180 − 66.67 = 113.33°. |
| 5. Why tan 210° is positive | B. cos = −0.866 and sin = −0.5, so the ratio is positive. |
Questions 1 and 2 are the same reflection asked twice on purpose. A student who gets 1 right and 2 wrong has learnt a rule rather than a picture, and that is exactly the diagnosis worth having.
1 markThe acute angle from the inverse function.
1 markThe second angle, 180° minus it.
1 markTesting each against the angle sum.
1 markBoth answers, or a stated reason for rejecting one.
The third mark is the one almost nobody writes. Insist on the line "180 − 40 − 113.33 = 26.67 > 0, so this one exists". It takes six seconds and it is a mark, and it is also the only thing that stops a student offering a second triangle when there is not one.
| They wrote | What happened |
|---|---|
| −0.866 for sin 120° | The error this page exists for, inverted. They have applied the cosine's behaviour to the sine. Send them to the figure and ask which leg is the height. |
| +0.5 for cos 120° | The same error the other way: the size is right and the sign was not changed. Ask which side of the y-axis the point is on. |
| 0.5 for sin 120° | Cosine and sine swapped. Common, and worth naming: across first, up second, like coordinates. |
| −0.5774 for tan 210° | The commonest answer on question 3. They assumed the third quadrant makes everything negative. Two minuses cancel. This is why the tangent row of the sign table should be derived and never memorised. |
| 1.7321 for tan 210° | Used 60° as the acute partner instead of 30°. 210 − 180 = 30, and the symmetry is about the x-axis, not the diagonal. |
| 23.33 on question 4 | 90 − 66.67. They have reflected in the wrong line. The sine repeats about the y-axis, so the partner is 180 − B. |
| 73.33 on question 4 | That is the third ANGLE of the acute triangle. They have answered a question one step further on, which usually means they worked the whole thing and lost track of what was asked. |
| One third side only | They trusted the calculator. The machine cannot return two values from one inverse sine, and that is a property of functions, not a limitation to work around. |
"Why is tan 90° undefined and not infinity?" Because the sine is 1 and the cosine is 0, and dividing by zero is not an operation. The graph goes off to infinity on one side and comes back from minus infinity on the other, so there is no single value to assign, not even a large one. Saying "infinity" sounds like an answer and loses the mark; "undefined" is the answer.
"Do I need the exact values if I have a calculator?" On Paper 1 there is no calculator, and the question will be built so the exact form exists. √3/2 earns the mark and 0.866 does not, because 0.866 is a rounding of an exact number the question wanted. It is also faster once learnt, and the two triangles behind it mean there are five values, not fifteen.
"Which is the acute partner?" Whichever acute angle the point reflects onto. For 120° it is 60°; for 210° it is 30°; for 300° it is 60°. The reliable instruction is the distance to the nearest point on the x-axis, which is 0°, 180° or 360°, never 90°. Students who measure to 90° get 30° and 60° the wrong way round, and it is worth saying out loud.
"When is there only one triangle?" When the obtuse option leaves nothing for the third angle. Two examples are worth the time, and the first is the one that surprises people. Give them A = 100°, a = 12, b = 10: sin B = 0.8207, so B is 55.15° or 124.85°, and 100 + 124.85 = 224.85, which is past 180°. One triangle. Then give them A = 100°, a = 9.9, b = 10, where sin B = 0.9948, comfortably under 1 and perfectly legal, the calculator returns 84.13° without complaint, and there is no triangle at all: 100 + 84.13 = 184.13 and 100 + 95.87 = 195.87, so both options fail. That example is the useful one because it shows sin B < 1 is not enough. The test is always the angle sum.
If you want a case where the sine itself is impossible, A = 50°, a = 5, b = 10 gives sin B = 1.5321 and the calculator errors. Useful, but less instructive than the 9.9 one, because the machine does the rejecting for them.
| Demonstrate | Put sin(60°) and sin(120°) on consecutive lines, then cos(60°) and cos(120°). The sine pair agree and the cosine pair do not. Four entries, thirty seconds, and the misconception has nowhere to hide. Do it before the explanation rather than after. |
|---|---|
| Where they stick | Believing the machine refused to give the second angle because they typed something wrong. It did not: sin⁻¹ is a function and returns one value by design. Say that plainly, because students who think it is a setting go looking for it. |
| The check | Feed the second angle back in. sin(113.33°) returns 0.9183, the value they started from, which proves the second solution rather than asserting it. |
Set the angle unit deliberately at the start of the lesson and say why: Analysis assumes radians unless a degree symbol appears, and this sub-topic is the one where the two notations sit side by side.
| Step | What |
|---|---|
| 1 | The unit circle and the definition. Cosine across, sine up, read as coordinates. |
| 2 | Why neither leaves −1 to 1, from the radius. |
| 3 | tan as the gradient of the line through the origin. The guide's y = x tan θ. |
| 4 | The two triangles, half a square and half an equilateral, and the five exact values. |
| 5 | The figure at 60°, then the mirror. Collect what moved. |
| 6 | Derive the sign table from the picture rather than giving it. The tangent row last, as a product. |
| 7 | tan 210°, which catches the everything-is-negative assumption. |
| 8 | The ambiguous case, with both third sides worked out: 10.43 and 4.89. |
Do not teach CAST or ASTC as the primary tool. The mnemonics are fine as a check and useless as an explanation, and a student who has only the mnemonic cannot tell you why the tangent is positive in the third quadrant. The circle answers that in one line and the mnemonic never does. If they already know a mnemonic, let them keep it and insist they can also derive it.
Do not say "remember the sine rule can have two answers". Remembering it works until the one question where there is only one answer, and then they invent a second triangle and lose the mark for it. Say instead: take the inverse sine, write 180 minus it beside the answer, and test both against the angle sum. That procedure gives one answer when there is one and two when there are two, which remembering cannot do.