Special Triangles and Exact Values

Take a square and slice it along a diagonal. Both legs are equal (say length $1$), and by the Pythagorean theorem the hypotenuse is $\sqrt{1^2+1^2}=\sqrt{2}$. The two acute angles must be equal because the legs are equal (an isosceles right triangle), and since they share the remaining $90°$, each is $45°$. Reading the ratios straight off the sides: $\sin 45°=\cos 45°=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$ (multiply top and bottom by $\sqrt{2}$ to rationalize), and $\tan 45°=\frac{1}{1}=1$. The fact that $\tan 45°=1$ is worth remembering on its own: a $45°$ slope rises exactly as much as it runs.

Take an equilateral triangle with side length $2$ and slice it in half through one vertex. Every angle of an equilateral triangle is $60°$, and the cut splits the top angle into two $30°$ angles while meeting the base at $90°$. The half-base is $1$, the full side (still $2$) becomes the hypotenuse, and the height is $\sqrt{2^2-1^2}=\sqrt{3}$ by the Pythagorean theorem. This gives a $30°$-$60°$-$90°$ triangle with sides $1$, $\sqrt{3}$, and $2$. The shortest side sits opposite the smallest angle ($30°$), and it is exactly HALF the hypotenuse, which is the fastest fact to check on a sketch.

Look for the patterns rather than memorizing 15 separate entries. Going down the table, sine climbs from $0$ to $1$ while cosine falls from $1$ to $0$: as the angle opens up, the opposite side grows and the adjacent side shrinks. The sine and cosine columns are mirror images because $30°$ and $60°$ are complementary: the side opposite $30°$ is the side adjacent to $60°$, so $\sin 30°=\cos 60°$ and $\sin 60°=\cos 30°$. Tangent is always sine divided by cosine, so $\tan 60°=\frac{\sqrt{3}/2}{1/2}=\sqrt{3}$. The reciprocal ratios come free by flipping each entry, as in the table below.

Trigonometry - Special triangles

Picture a circle of radius $1$ centred at the origin. For any angle $\theta$ measured counterclockwise from the positive $x$-axis, the point where the angle's arm crosses the circle has coordinates $(\cos \theta ,\sin \theta )$. Why? Drop a perpendicular to the $x$-axis: you get a right triangle with hypotenuse $1$, so $\cos \theta =\frac{x}{1}=x$ and $\sin \theta =\frac{y}{1}=y$. That's really the deeper definition of sine and cosine, and it's what lets these ratios extend to angles beyond $0°$–$90°$ (the next topic). It also explains the $0°$ and $90°$ rows: at $0°$ the point is $(1,0)$, so $\cos 0°=1$ and $\sin 0°=0$; at $90°$ it is $(0,1)$, so $\cos 90°=0$, $\sin 90°=1$, and $\tan 90°=\frac{1}{0}$ is undefined. MCR3U works entirely in degrees; the "radian" unit you may see elsewhere is just a different way to measure the same angle and isn't used in this course.

Exact-value questions are algebra with surds. Substitute each exact value in brackets, then simplify with the usual rules: $\sqrt{a}\cdot \sqrt{a}=a$, like surds add ($\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}=\sqrt{3}$), and a surd in a denominator can be rationalized by multiplying top and bottom by it. Watch the notation ${\sin }^2\theta$: it means ${(\sin \theta )}^2$, so ${\sin }^{2}60°={\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{3}{4}$. Special triangles also solve geometry problems exactly: if a $60°$ angle of elevation is measured $20$ m from a tower, the height is $20\tan 60°=20\sqrt{3}$ m, with no rounding at all. Recognize the method from the angles: if every angle in a problem is $30°$, $45°$, $60°$ or a multiple of $90°$, and the question says exact, reach for the special triangles instead of the calculator.