Home page for accesible maths Math 101 Chapter 2: Functions of a real variable

Style control - access keys in brackets

Font (2 3) - + Letter spacing (4 5) - + Word spacing (6 7) - + Line spacing (8 9) - +

2.40 Appendix: Basic right-angled triangle

Pythagoras’ Theorem

cos2⁡x+sin2⁡x=1\cos^{2}x+\sin^{2}x=1

Let the right-angled triangle have angle xx, opposite BB, hypotenuse HH, and adjacent AA. (It is a bad choice of notation to use OO for opposite.)

(i) Let A=0A=0 and x=0x=0, then sin⁡0=0,\sin 0=0, cos⁡0=1\cos 0=1 and tan⁡0=0.\tan 0=0.

(ii) Let B=0B=0 and x=π/2x=\pi/2, then sin⁡π2=1,\sin{{\pi}\over{2}}=1, cos⁡π2=0\cos{{\pi}\over{2}}=0 and tan⁡x→∞\tan x\rightarrow\infty as x→π2-x\rightarrow{{\pi}\over{2}}-.

(iii) Let B=A=1B=A=1 and H=2H=\sqrt{2} and x=π4x={{\pi}\over{4}}, then sin⁡π4=cos⁡π4=12\sin{{\pi}\over{4}}=\cos{{\pi}\over{4}}={{1}\over{\sqrt{2}}}, and tan⁡π4=1\tan{{\pi}\over{4}}=1.

(iv) Let B=1,B=1, H=2H=2 and A=3A=\sqrt{3}, and x=π6x={{\pi}\over{6}}, then sin⁡π6=12,\sin{{\pi}\over{6}}={{1}\over{2}}, cos⁡π6=32\cos{{\pi}\over{6}}={{\sqrt{3}}\over{2}} and tan⁡π6=13\tan{{\pi}\over{6}}={{1}\over{\sqrt{3}}}.

Instead of measuring triangles, computers calculate trig functions by using the Maclaurin series of 4.1 below.