E3 - Trigonometric graphs and exact values

Exact values from two triangles

How can two little triangles tell you the exact value of $\sin(17\pi/3)$, an angle almost three full turns from zero? Start with a $45^\circ$ right triangle whose perpendicular sides are $1,1$: Pythagoras gives hypotenuse $\sqrt2$, so its sine and cosine are both $\sqrt2/2$.

Bisect an equilateral triangle of side $2$. The half-base is $1$, the height is $\sqrt3$ and the hypotenuse is $2$. The resulting $30^\circ$ and $60^\circ$ ratios, together with the isosceles right triangle, generate the standard exact values.

Angles on the axes need unit-circle coordinates instead of a flattened triangle: $0$ gives $(1,0)$, $\pi/2$ gives $(0,1)$, and $\pi$ gives $(-1,0)$.

The table records what the triangles and axis points give. These exact values are not in the formula booklet: learn them.

Tangent's undefined entry is division by zero, not a very large number. Every odd multiple of $\pi/2$ has the same problem; every integer multiple of $\pi$ has tangent zero.

The three acute sine squares have a neat balance: $1/4+1/2+3/4=3/2$. The corresponding cosine squares give the same sum in reverse order. The symmetry between the two triangles makes these different-looking surds fit together exactly.

The following question distinguishes an angle from a function value.

Signs, symmetry and repeated turns

A terminal radius at $7\pi/6$ lies in quadrant III, just $\pi/6$ past a half-turn. Its vertical coordinate is negative, so its sine is $-1/2$.

The reference angle is the acute angle between a non-axis terminal radius and the nearest horizontal axis. Use it to find a magnitude, then use the quadrant to find the sign. At an axis, use coordinates directly.

A period is a positive horizontal shift reproducing every value and the domain. The smallest positive period is $2\pi$ for sine and cosine, and $\pi$ for tangent. A half-turn changes both coordinate signs, so it preserves their quotient but not either coordinate individually.

Supplementary angles have equal sine, opposite cosine and opposite tangent where defined. Thus $\cos(11\pi/6)=\sqrt3/2$, while $\tan(-5\pi/4)=\tan(-\pi/4)=-1$. These relations are not supplied in the booklet: learn them from the circle.

Change only the number of full turns by adding $2\pi$. The terminal point does not move.

The same process answers the opening question: $17\pi/3-4\pi=5\pi/3$, which has negative sine and reference angle $\pi/3$. Therefore $\sin(17\pi/3)=-\sqrt3/2$. A large-looking angle has become a small-triangle ratio.

Add one more full turn in the matched question.

Sine and cosine graph features

Recording the rotating unit-circle point's height against its angle gives the sine wave; recording its horizontal coordinate gives cosine.

Both graphs have domain $\mathbb R$, range $[-1,1]$ and period $2\pi$, or $360^\circ$ on a degree axis. Over one radian turn, sine's values at $0,\pi/2,\pi,3\pi/2,2\pi$ are $0,1,0,-1,0$; cosine's are $1,0,-1,0,1$.

Sine has half-turn symmetry about the origin; cosine has reflection symmetry in the vertical axis. Cosine is sine shifted left by $\pi/2$, so $\cos x=\sin(x+\pi/2)$. These graph facts and the phase relationship are not supplied in the formula booklet. The supplied addition formula can also recover the phase relationship.