From graphs to equations You can already read values off the sine, cosine and tangent graphs. Solving an equation such as \(\sin x=k\) runs that skill in reverse: find every angle in a stated interval that produces the value \(k\). The calculator returns only one angle — the principal value: \(\sin^{-1}\) gives an angle from \(-90^\circ\) to \(90^\circ\), \(\cos^{-1}\) from \(0^\circ\) to \(180^\circ\) and \(\tan^{-1}\) between \(-90^\circ\) and \(90^\circ\) — so the remaining solutions must be built by hand. The reference angle For sine or cosine, first require \(|k|\leq1\). For a non-axis solution, use the non-negative reference angle \(\alpha=\sin^{-1}|k|\), \(\cos^{-1}|k|\) or \(\tan^{-1}|k|\), in the same units as the interval. Handle axis values separately: a reference angle may be \(0^\circ\) or \(90^\circ\), so it is not always acute. Quadrants from the sign of \(k\) (ASTC) QuadrantRatio positive thereSolution (degrees)Solution (radians) I II III IV Method: Ratio, Reference angle, Quadrants, Angles Rearrange the equation so a single ratio stands alone on one side. Find the reference angle \(\alpha\) from the positive value, in the same mode (degrees or radians) as the interval. Use the sign of \(k\) to choose the quadrants, then write the angle in each chosen quadrant. List every solution that lies inside the stated interval. Reading solutions from a graph In the illustrated case \(k=\frac12\), the horizontal line cuts one full sine cycle twice. Thus \(\sin x=\frac12\) has two solutions in \(0\leq x\leq2\pi\): \(x=\alpha\) and \(x=\pi-\alpha\). The count is different at extrema or when both endpoints are included for \(k=0\). Special values: read from the graph \(y=\sin x\), \(0\le x\le 2\pi\): peak at \(\dfrac{\pi}{2}\), zeros at \(0\), \(\pi\), \(2\pi\), trough at \(\dfrac{3\pi}{2}\). \(y=\cos x\), \(0\le x\le 2\pi\): peaks at \(0\) and \(2\pi\), zeros at \(\dfrac{\pi}{2}\), \(\dfrac{3\pi}{2}\), trough at \(\pi\). EquationSolutions for \(0^\circ\le x\le 360^\circ\)Solutions for \(0\le x\le 2\pi\) \(\sin x=1\) \(\sin x=0\) \(\sin x=-1\) \(\cos x=1\) \(\cos x=0\) \(\cos x=-1\) Exam check Set the calculator mode to match the interval — degrees for \(0^\circ\le x\le 360^\circ\), radians for \(0\le x\le 2\pi\). Never round the reference angle mid-working; round only the final answers.

Reference Angles and the Quadrant Method

Secondary 4

TGM Original Questions

From graphs to equations

You can already read values off the sine, cosine and tangent graphs. Solving an equation such as \(\sin x=k\) runs that skill in reverse: find every angle in a stated interval that produces the value \(k\). The calculator returns only one angle — the principal value: \(\sin^{-1}\) gives an angle from \(-90^\circ\) to \(90^\circ\), \(\cos^{-1}\) from \(0^\circ\) to \(180^\circ\) and \(\tan^{-1}\) between \(-90^\circ\) and \(90^\circ\) — so the remaining solutions must be built by hand.

The reference angle

For sine or cosine, first require \(|k|\leq1\). For a non-axis solution, use the non-negative reference angle \(\alpha=\sin^{-1}|k|\), \(\cos^{-1}|k|\) or \(\tan^{-1}|k|\), in the same units as the interval. Handle axis values separately: a reference angle may be \(0^\circ\) or \(90^\circ\), so it is not always acute.

Quadrants from the sign of \(k\) (ASTC)

Quadrant	Ratio positive there	Solution (degrees)	Solution (radians)
I
II
III
IV

Method: Ratio, Reference angle, Quadrants, Angles

Rearrange the equation so a single ratio stands alone on one side.
Find the reference angle \(\alpha\) from the positive value, in the same mode (degrees or radians) as the interval.
Use the sign of \(k\) to choose the quadrants, then write the angle in each chosen quadrant.
List every solution that lies inside the stated interval.

Reading solutions from a graph

In the illustrated case \(k=\frac12\), the horizontal line cuts one full sine cycle twice. Thus \(\sin x=\frac12\) has two solutions in \(0\leq x\leq2\pi\): \(x=\alpha\) and \(x=\pi-\alpha\). The count is different at extrema or when both endpoints are included for \(k=0\).

Special values: read from the graph

\(y=\sin x\), \(0\le x\le 2\pi\): peak at \(\dfrac{\pi}{2}\), zeros at \(0\), \(\pi\), \(2\pi\), trough at \(\dfrac{3\pi}{2}\).

\(y=\cos x\), \(0\le x\le 2\pi\): peaks at \(0\) and \(2\pi\), zeros at \(\dfrac{\pi}{2}\), \(\dfrac{3\pi}{2}\), trough at \(\pi\).

Equation	Solutions for \(0^\circ\le x\le 360^\circ\)	Solutions for \(0\le x\le 2\pi\)
\(\sin x=1\)
\(\sin x=0\)
\(\sin x=-1\)
\(\cos x=1\)
\(\cos x=0\)
\(\cos x=-1\)

Exam check

Set the calculator mode to match the interval — degrees for \(0^\circ\le x\le 360^\circ\), radians for \(0\le x\le 2\pi\). Never round the reference angle mid-working; round only the final answers.

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