A trigonometric equation may be easy to solve on 0° to 360° but much harder when the given interval begins at a negative angle and crosses zero. The reliable method is to find the general solution family, shift those solutions by whole periods, and keep only the angles inside the stated interval.
For example, sin θ = 1/2 for −360° ≤ θ ≤ 0° has exactly two solutions: −330° and −210°. The ordinary positive angles 30° and 150° are not in the requested interval, but subtracting 360° from each produces the correct negative solutions. This guide covers sine, cosine, tangent, double-angle arguments, shifted angles, endpoint inclusion and degree–radian consistency in Additional Mathematics.
Negative angles follow the same unit circle
An angle measured clockwise from the positive x-axis is often described as negative, while anticlockwise angles are positive. Both represent real rotation and can be used in trigonometric functions.
The unit circle repeats its position after each full revolution. Thus sin(θ − 360°) = sin θ and cos(θ − 360°) = cos θ, while tangent repeats every 180° where defined.
The important implication is that a negative solution can be obtained by adding or subtracting whole periods from a familiar positive solution. The sign of the angle does not decide whether its sine or cosine is positive.
A learner who rejects every negative angle because it looks unfamiliar will miss valid answers when an interval includes clockwise rotations. The given interval, not personal preference for positive reference angles, determines which representatives are required.
Read the interval before solving
An equation and an interval form one complete problem. Before calculating a reference angle, identify the lower and upper endpoints and whether each is included.
For example, −360° ≤ θ ≤ 0° includes both −360° and 0°. By contrast, −360° < θ < 0° excludes them. This distinction may matter for equations whose solutions occur exactly at those boundaries.
Write the interval beside the equation and keep it visible. It is particularly easy to lose it when an intermediate substitution introduces a new angle variable or a double-angle expression.
When the problem uses radians, preserve radians throughout rather than mixing 180° and π without converting. Consistent angle units are an essential condition for correct solution listing.
Build the general solution for sine
For sin θ = 1/2, the familiar positive solutions in one revolution are 30° and 150°. Since sine has period 360°, the complete degree-mode solution families are θ = 30° + 360°k and θ = 150° + 360°k, where k is any integer.
This representation does more than give one pair of answers. It gives every angle whose sine is 1/2, including values outside the usual 0° to 360° interval.
To solve a restricted-interval question, choose only those integers k for which the resulting angle lies between the stated endpoints.
The student need not memorise dozens of negative-angle values. The general solution already describes them. Finding all solutions is a filtering task built on periodicity and the correct reference-angle families.
Worked example: −360° to 0°
Solve sin θ = 1/2 for −360° ≤ θ ≤ 0°. Start with θ = 30° + 360°k or θ = 150° + 360°k.
For k = −1, these give θ = −330° and θ = −210°. Both are inside the requested interval.
For k = 0, they give 30° and 150°, which lie above the upper endpoint and must be discarded. For k = −2, they give −690° and −570°, below the lower endpoint.
Therefore the complete solution set is θ = −330° or −210°. Verify using periodicity: sin(−330°) = sin30° = 1/2 and sin(−210°) = sin150° = 1/2.
This is a more dependable explanation than reading calculator inverse-sine output and guessing that one negative angle should be appended.
Why the ordinary 30° answer is not automatically acceptable
A calculator may report 30° for sin⁻¹(1/2), depending on angle mode and principal-value convention. That is one valid angle but not necessarily an angle in the question’s interval.
For −360° ≤ θ ≤ 0°, 30° is too large, so it is not an acceptable final solution. Subtracting 360° gives the coterminal angle −330°, which belongs.
Likewise, the second family at 150° produces −210° after subtracting 360°. Both must be included.
The calculator’s principal value is a starting point for reference-angle reasoning, not a substitute for interval filtering. Students should distinguish a valid equation solution in the unrestricted sense from a valid answer to the restricted problem.
The interval −180° to 180° produces a different list
Solve the same equation sin θ = 1/2 for −180° ≤ θ ≤ 180°. The two familiar angles 30° and 150° already lie in this interval.
The coterminal negative versions −330° and −210° are below −180° and therefore excluded. There are no other permitted solutions.
Thus the answer is θ = 30° or 150°. The equation has not changed; the allowed interval has.
This comparison demonstrates why students should read the actual lower and upper endpoints before listing solutions. There is no universal instruction to “include negative versions” or “keep positive versions”; only the interval decides.
When sine is negative on a symmetric interval
Solve sin θ = −1/2 for −180° ≤ θ ≤ 180°. The reference angle is 30°. On this symmetric interval, the valid negative angles are −150° and −30°.
Check: sin(−150°) = −1/2 and sin(−30°) = −1/2. Both lie inside the interval.
The familiar positive solutions 210° and 330° from the interval 0° to 360° are not permitted because they exceed 180°. Subtracting 360° gives −150° and −30°, respectively.
This is a useful way to practise the sign of sine. A negative function value does not mean that every associated angle must be negative on every possible interval, but for this particular symmetric interval the allowed solutions happen to be negative.
Cosine solutions can appear as a positive and negative pair
Solve cos θ = −1/2 for −180° ≤ θ ≤ 180°. The familiar angles in a standard revolution are 120° and 240°.
The angle 120° lies inside the interval. The coterminal angle 240° − 360° = −120° also lies inside. Therefore the solutions are θ = −120° and 120°.
This result reflects the evenness of cosine: cos(−θ) = cos θ. The two angles have the same cosine because their unit-circle positions are reflected across the x-axis.
Use this symmetry to check the result, but still solve the interval question exactly. A wider interval could contain more angles separated by full periods.
Tangent uses a period of 180°
The tangent function repeats every 180° in degree measure. Thus tan(θ + 180°) = tan θ wherever tangent is defined. Its general solution uses steps of 180°, not 360° alone.
For tan θ = 1, one solution is 45°, so the complete degree-mode family is θ = 45° + 180°k, with integer k.
On −180° ≤ θ ≤ 180°, choose k = −1 to get −135° and k = 0 to get 45°. The next values, −315° and 225°, are outside the interval.
Therefore the answer is −135° or 45°. A learner who shifts tangent solutions only by 360° would miss one of these angles because tangent has a shorter period.
A tangent example with negative value
For tan θ = −1, the standard reference-angle solutions include 135° and 315°. Using the 180° period, one general family is θ = 135° + 180°k.
On −180° ≤ θ ≤ 180°, k = −1 gives θ = −45°, while k = 0 gives 135°. These are the two permitted solutions.
A check with the oddness of tangent gives tan(−45°) = −tan45° = −1. The value at 135° is also −1 because sine is positive and cosine is negative in the second quadrant.
This reinforces the importance of choosing the correct fundamental period. Different trigonometric functions use different repeating intervals, so their solution families should not be copied from one another mechanically.
A wide interval can contain more than two sine solutions
Solve sin θ = 1/2 on −360° ≤ θ ≤ 360°. The two solution families are 30° + 360°k and 150° + 360°k.
For k = −1, they give −330° and −210°. For k = 0, they give 30° and 150°. For k = 1, they give 390° and 510°, both above 360°.
Therefore there are four permitted solutions: −330°, −210°, 30° and 150°. The interval spans two full revolutions, so seeing four rather than two solutions is reasonable.
Do not assume a trigonometric equation has a fixed number of answers independently of its interval. Wider intervals can contain multiple representatives of each periodic family.
Shifted intervals need deliberate endpoint filtering
Consider sin θ = 1/2 for −90° ≤ θ ≤ 450°. The general families remain 30° + 360°k and 150° + 360°k.
The values 30° and 150° lie inside the interval. Adding 360° to the first gives 390°, also inside; adding 360° to the second gives 510°, too large. Subtracting 360° gives −330° and −210°, both below −90°.
The complete solution set is therefore 30°, 150° and 390°. The interval contains one and a half revolutions, and it captures an extra representative of only one family.
This example shows why students should calculate bounds for each family rather than assume every pair of solutions repeats together within an arbitrary interval.
Inclusive endpoints can themselves be solutions
Solve sin θ = 0 for −180° ≤ θ ≤ 180°. The general family is θ = 180°k for integer k.
Within the inclusive interval, the permitted values are −180°, 0° and 180°. The two endpoint values must both be listed because sine vanishes there.
If the interval were −180° < θ < 180°, only 0° would remain. The equation is unchanged, but the strict inequalities exclude the endpoints.
This is a common examination trap. A student who rounds or lists interior angles correctly can still lose completeness by overlooking equality signs at the boundaries.
Convert a double-angle equation into an interval for the new variable
Solve sin(2θ) = 1/2 for −180° ≤ θ ≤ 180°. Let φ = 2θ. The allowed interval for φ is then −360° ≤ φ ≤ 360°.
Solve sin φ = 1/2 on that doubled interval. The permitted φ-values are −330°, −210°, 30° and 150°.
Now divide each by two to recover θ: −165°, −105°, 15° and 75°. All four lie in the original interval −180° to 180°.
The interval must be doubled before solving, not after listing only 30° and 150°. If the student applies the ordinary one-revolution answer directly and halves it, the two negative-angle solutions are lost.
Why dividing the argument changes the number of solutions
The equation sin(2θ) = 1/2 oscillates twice as fast as sin θ = 1/2 when θ moves through a fixed angular interval. Over −180° to 180°, the inner angle 2θ covers −360° to 360°.
That wider inner interval contains more repetitions of the sine pattern, so four values of θ satisfy the equation rather than only two.
This is a conceptual explanation for why a double-angle equation often produces more roots on the same stated θ-interval. The number is determined by the inner argument’s range and the trigonometric period.
Students should not simply double the reference angle or double the number of answers without checking. Transform the interval, solve the new variable and map every valid solution back.
An angle shift changes the interval for the inner argument
Consider sin(θ + 60°) = 1/2 on −180° ≤ θ ≤ 180°. Let φ = θ + 60°. The inner interval becomes −120° ≤ φ ≤ 240°.
The sine equation gives φ = 30° or 150° within that interval. The negative coterminal versions −330° and −210° lie below −120°, so they are excluded.
Subtract 60° to recover θ. The solutions are θ = −30° and 90°. Check: when θ = −30°, the argument is 30°; when θ = 90°, the argument is 150°.
This is safer than applying familiar sine-angle solutions directly to θ and forgetting that the sine function is acting on θ + 60°, not θ alone.
Combining a shift with a doubled argument
Suppose the equation is sin(2θ + 30°) = 1/2 for −180° ≤ θ ≤ 180°. Define φ = 2θ + 30°. The inner interval is −330° ≤ φ ≤ 390°.
The permitted sine solutions are φ = −330°, −210°, 30°, 150° and 390°. The endpoints −330° and 390° are both included because the original θ bounds are inclusive.
Solve 2θ + 30° = φ for each permitted value. This gives θ = −180°, −120°, 0°, 60° and 180°.
Direct substitution verifies the endpoints: at θ = −180° the argument is −330°, with sine 1/2; at θ = 180° the argument is 390°, also with sine 1/2. The full solution list has five values, so assuming a fixed pair or quartet would be wrong.
Radians follow the same periodic reasoning
In radians, sine and cosine repeat every 2π, while tangent repeats every π. The unit-circle logic is unchanged; only the numerical unit of angle differs.
For sin θ = 1/2 on −π ≤ θ ≤ π, the permitted solutions are π/6 and 5π/6. Their negative coterminal counterparts −11π/6 and −7π/6 lie outside the specified interval.
For sin θ = −1/2 on the same interval, the solutions are −5π/6 and −π/6. The reference angles and signs agree with the degree-mode versions.
Do not mix degree values such as 30° with radian intervals such as [−π, π] without converting. Before using a calculator, check whether its angular mode matches the question.
A calculator gives a reference value, not the complete set
A calculator’s inverse trigonometric function usually reports a principal value from a limited interval. It does not automatically list all coterminal solutions in the user’s question.
For sin θ = 1/2, the calculator may report 30°. The second sine family at 150° still needs quadrant reasoning, and both families need shifts by 360° when the allowed interval extends into negative or larger angles.
For tan θ = 1, the principal 45° belongs to a family repeating every 180°. That period is different from sine’s and cosine’s.
A good examination routine uses the calculator to identify a reference angle where appropriate, then uses unit-circle signs, periods and stated endpoints to construct the final exact set.
Use integer bounds to guarantee completeness
A systematic method is to take each general solution family and solve an inequality for its integer parameter k. For sin θ = 1/2, one family is θ = 30° + 360°k.
If the interval is −360° ≤ θ ≤ 0°, then −360 ≤ 30 + 360k ≤ 0. Subtract 30 and divide by 360 to obtain −390/360 ≤ k ≤ −30/360. The only integer k in this interval is −1.
For the second family θ = 150° + 360°k, the analogous bounds again permit only k = −1. Hence the solutions are −330° and −210°.
This parameter-bound method is especially useful for unusual intervals or longer trigonometric periods. It gives evidence that no solutions were missed.
A simple solution table prevents omissions
For a trigonometric equation with multiple families, make a small table of base solutions, period shifts and candidate angles in the specified interval.
For sin θ = 1/2 on −360° to 360°, the family based at 30° contributes −330° and 30°. The family based at 150° contributes −210° and 150°. Combine and order the four values.
For tan θ = 1 on −180° to 180°, only one family is needed, based at 45° with 180° shifts, giving −135° and 45°.
A table is not compulsory, but it makes the search inspectable. The learner can see which family produced each angle and which integer shifts have been tested.
Why signs alone cannot determine the answer list
A negative angle does not necessarily have negative sine, cosine or tangent. For example, sin(−330°) = 1/2 is positive, cos(−120°) = −1/2 is negative, and tan(−135°) = 1 is positive.
These values are governed by positions on the unit circle and by periodicity, not by the printed minus sign in front of the angle.
A student who reasons “the question wants sine positive, so θ must be positive” is using an invalid shortcut. A clockwise rotation can end at the same point as a positive coterminal angle.
The correct method uses the sign of the trigonometric ratio in the appropriate quadrants, then chooses angle representatives that fit the given interval.
What changes when the interval is open?
Suppose an angle equation has a solution exactly at θ = −180° or θ = 180°. Whether that value is included depends on the interval’s strictness.
For sin θ = 0, the interval −180° ≤ θ ≤ 180° gives −180°, 0° and 180°. The interval −180° < θ < 180° gives only 0°.
For sin θ = 1/2 on the same pair of intervals, neither endpoint is a solution, so the answer remains 30° and 150° in both cases. Thus changing endpoint inclusion does not always change the output set.
Students should check boundary values only when they might satisfy the equation. The general-solution families plus inequality filtering determine this exactly.
How to detect a missed root before submitting
After listing solutions, put them in numerical order and check their location in the specified interval. Then inspect the gap between consecutive solutions and the relevant period to ask whether another coterminal value could lie in between.
For sin θ = 1/2 on −360° to 360°, the complete list is −330°, −210°, 30° and 150°. The pattern alternates gaps of 120° and 240° because the two sine families are not uniformly separated.
If only two angles are listed over two full revolutions, review both families and their period shifts. If an angle lies outside the interval, discard it even when direct substitution makes the equation true.
Completeness has two parts: every reported solution must be valid, and every valid solution in the interval must be reported. Both need attention.
Six practice questions with checked answers
Question one: sin θ = 1/2 on −360° ≤ θ ≤ 0°. Answer: −330°, −210°.
Question two: sin θ = −1/2 on −180° ≤ θ ≤ 180°. Answer: −150°, −30°.
Question three: cos θ = −1/2 on −180° ≤ θ ≤ 180°. Answer: −120°, 120°.
Question four: tan θ = 1 on −180° ≤ θ ≤ 180°. Answer: −135°, 45°.
Question five: sin(2θ) = 1/2 on −180° ≤ θ ≤ 180°. Answer: −165°, −105°, 15°, 75°.
Question six: sin(θ + 60°) = 1/2 on −180° ≤ θ ≤ 180°. Answer: −30°, 90°.
For every question, write the general family and filter its integer period shifts. Then verify at least one negative-angle solution with periodicity or exact trigonometric values.
A concise examination routine
First identify the trigonometric function and its period. Second read the exact angle interval and its endpoint signs. Third solve for the general angle family using reference-angle reasoning.
Fourth, if the function acts on 2θ, θ + c or another expression, translate the permitted interval for that entire argument. Fifth list the integer period shifts that fall within that interval.
Sixth map solutions back to θ, discard values outside the original interval, and order the remaining angles. Seventh check boundary values and the calculator’s degree or radian mode.
This routine replaces random angle guessing with a short, logically complete process. It is especially helpful when the interval crosses zero and the familiar positive angles no longer fit.
Singapore Additional Mathematics context and further study
Trigonometric functions, identities and equations are central mathematical ideas, but the exact range of equation types and angle conventions must match the learner’s school and examination syllabus.
For Singapore, check the relevant SEAB lists for 2026 O-Level Additional Mathematics 4049, 2027 SEC G3 K341 or 2027 SEC G2 K232. Use syllabus-appropriate variations and teacher instructions.
Related articles include How to Improve Additional Mathematics | Finding Every Trigonometric Solution, How to Distinguish Horizontal and Vertical Graph Shifts and How to Build a Weekly A-Math Question Queue.
For more learning resources, visit the Additional Mathematics archive or enquire about eduKate Singapore three-pax tutorials near Sixth Avenue MRT. The goal is not simply remembering a pair of angles: it is proving that every valid representative in the stated interval has been found. Properly taught kids shine a bright light into the future.