Numbers & Functions · 6 Trigonometric Functions

6.3 The special angles (built from two triangles)

To compute with sine and cosine you need their values at a handful of angles: π 6 , π 4 , π 3 (that’s 30◦, 45◦, 60◦). Most books make you memorize a table. We won’t — because all six values come from just two triangles you can draw any time you forget them.

1 1 √ 2 45◦ cos π 4 = sin π 4 = 1 √ 2 √ 3 1 2 30◦ 60◦ cos π 3 = 1 2, sin π 3 = √ 3 2 Figure 6.4. The two triangles behind every special angle. Half a unit square gives the 45◦values; half an equilateral triangle (side 2) gives the 30◦and 60◦values. Pythagoras supplies the √ 2 and √ 3. The picture is the memory.

Triangle one — 45◦, from half a square. Cut a unit square along its diagonal. Two equal short sides (both 1), two 45◦angles, and a diagonal of √ 12 + 12 = √ 2. Scale to hypotenuse 1 and read off cos π 4 = sin π 4 = 1 √ 2 ≈0.707.

They’re equal — at 45◦the walker is exactly diagonal.

Triangle two — 30◦and 60◦, from half an equilateral triangle. Take an equilateral triangle with all sides 2 and split it down the middle. Each half has hypotenuse 2, base 1, and height √ 22 −12 = √ 3. Scale to hypotenuse 1: cos π 3 = 1 2, sin π 3 = √ 3 2 ; cos π 6 = √ 3 2 , sin π 6 = 1 2.

Notice π 6 and π 3 have their sine and cosine swapped — they’re the two angles of the same triangle, seen from opposite corners.

6.4 Reading the rest off the walker: reduction identities

You’ll also need to relate the sine and cosine of one angle to a related angle — things like sin(π −θ). Again, no memorizing: it’s just “where does the walker land if you move the angle this way?”

And the complement swap: π 2 −θ reflects the walker across the 45◦diagonal, which swaps across and up — so sin(π 2 −θ) = cos θ. This is the deep reason cosine is literally “sine of the complement.” Don’t memorize the dozen reduction formulas — hold the walker in your mind and read the answer off the circle. Trigonometric Functions θ π −θ same height ⇒sin(π −θ) = sin θ opposite across ⇒cos(π −θ) = −cos θ Figure 6.5. A reduction identity is a reflection. The angle π −θ is the mirror of θ across the vertical axis — same height, opposite across-position — so sin(π −θ) = sin θ and cos(π −θ) = −cos θ, read straight off the circle.

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