Trigonometric Graphs and Ratios of Any Angle

Sketch y=\cos x for 0^\circ\leqslant x\leqslant 360^\circ and mark the point at x=40^\circ: the graph is symmetrical about x=180^\circ and about x=90^\circ (with the sign changing).

1. On the cos graph, \cos(180^\circ-40^\circ)=-\cos 40^\circ, so \cos 140^\circ=-k
2. The graph is symmetrical about x=180^\circ, so \cos 220^\circ=\cos 140^\circ=-k
3. \cos 320^\circ=\cos 40^\circ=k (positive, not negative)
4. \cos 130^\circ and \cos 50^\circ come from 50^\circ, not 40^\circ, so they are not \pm k
5. Answer: \cos 220^\circ

Picture the cos graph: where does it cross the x-axis between 0^\circ and 360^\circ, and how often does the pattern repeat?

1. For 0^\circ\leqslant x\leqslant 360^\circ the cos graph crosses the x-axis at 90^\circ and 270^\circ
2. The graph repeats every 360^\circ, so add 360^\circ to each: 450^\circ and 630^\circ
3. Answer: 90^\circ, 270^\circ, 450^\circ, 630^\circ

Decide which quadrant \theta is in from the signs of \sin\theta and \cos\theta before you use your calculator.

Part (a)

1. \sin^2\theta+\cos^2\theta=1, so \cos^2\theta=1-\dfrac{400}{841}=\dfrac{441}{841}
2. \cos\theta>0, so \cos\theta=\dfrac{21}{29}
3. \tan\theta=\dfrac{\sin\theta}{\cos\theta}=-\dfrac{20}{29}\div\dfrac{21}{29}
4. Answer: \tan\theta=-\dfrac{20}{21}

Part (b)

1. \sin\theta<0 and \cos\theta>0, so \theta is between 270^\circ and 360^\circ
2. The acute angle with \sin=\dfrac{20}{29} is \sin^{-1}\!\left(\dfrac{20}{29}\right)=43.60\ldots^\circ
3. Using the symmetry of the sine graph: \theta=360^\circ-43.60\ldots^\circ
4. Answer: \theta=316.4^\circ (1 d.p.)

Part (c)

1. -\tan\theta=\dfrac{20}{21}, which is positive
2. \tan^{-1}\!\left(\dfrac{20}{21}\right)=43.60\ldots^\circ
3. The tan graph repeats every 180^\circ, so the value between 180^\circ and 360^\circ is 180^\circ+43.60\ldots^\circ
4. Answer: x=223.6^\circ (1 d.p.)