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Concept 06 / 11Trigonometry

Trigonometric Ratios: Radian Measure, the Unit Circle, Signs and Allied Angles

Degree and radian measure with arc length and sector area, the six ratios defined on the unit circle, signs in the four quadrants (ASTC), values at standard angles, the Pythagorean identities, and reducing allied angles such as pi/2 +/- theta and pi +/- theta.

Figure 1
Radian measure. An angle of 11 radian subtends an arc equal to the radius. A full turn is 2π2\pi radians, so π\pi radians =180∘= 180^{\circ}, and 11 radian ≈57.3∘\approx 57.3^{\circ}. In radians, arc length is s=rθs = r\theta and sector area is 12r2θ\frac{1}{2}r^{2}\theta. Calculus always uses radians, because only then is ddxsin⁡x=cos⁡x\frac{d}{dx}\sin x = \cos x. The unit circle. Place an angle θ\theta at the origin, measured anticlockwise from the positive xx-axis. Its terminal side meets the unit circle at the point (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta), as in the figure. This defines sine and cosine for every real θ\theta, not just acute angles. The other ratios follow: tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta}, cot⁡θ=cos⁡θsin⁡θ\cot\theta = \frac{\cos\theta}{\sin\theta}, sec⁡θ=1cos⁡θ\sec\theta = \frac{1}{\cos\theta}, csc⁡θ=1sin⁡θ\csc\theta = \frac{1}{\sin\theta}, wherever the denominators are non-zero. For an acute angle in a right triangle these agree with 'opposite over hypotenuse' and so on. Pythagorean identities, from x2+y2=1x^{2} + y^{2} = 1: sin⁡2θ+cos⁡2θ=1\sin^{2}\theta + \cos^{2}\theta = 1; dividing by cos⁡2θ\cos^{2}\theta or sin⁡2θ\sin^{2}\theta: 1+tan⁡2θ=sec⁡2θ1 + \tan^{2}\theta = \sec^{2}\theta and 1+cot⁡2θ=csc⁡2θ1 + \cot^{2}\theta = \csc^{2}\theta. Useful consequence: sec⁡θ+tan⁡θ=1sec⁡θ−tan⁡θ\sec\theta + \tan\theta = \frac{1}{\sec\theta - \tan\theta}, since (sec⁡θ+tan⁡θ)(sec⁡θ−tan⁡θ)=1(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1. Ranges. sin⁡θ,cos⁡θ∈[−1,1]\sin\theta, \cos\theta \in [-1, 1]; sec⁡θ,csc⁡θ∈(−∞,−1]∪[1,∞)\sec\theta, \csc\theta \in (-\infty, -1] \cup [1, \infty); tan⁡θ,cot⁡θ\tan\theta, \cot\theta take all real values. Signs by quadrant. In quadrant I all ratios are positive; in II only sin⁡\sin (and csc⁡\csc); in III only tan⁡\tan (and cot⁡\cot); in IV only cos⁡\cos (and sec⁡\sec). The mnemonic 'All Students Take Coffee' (ASTC) follows the quadrants anticlockwise. Standard values: at 0,π6,π4,π3,π20, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}, sin⁡θ=02,12,22,32,42\sin\theta = \frac{\sqrt{0}}{2}, \frac{\sqrt{1}}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{4}}{2}, and cos⁡θ\cos\theta runs through the same list in reverse. Also worth knowing: sin⁡15∘=6−24\sin 15^{\circ} = \frac{\sqrt{6} - \sqrt{2}}{4}, tan⁡15∘=2−3\tan 15^{\circ} = 2 - \sqrt{3}, sin⁡18∘=5−14\sin 18^{\circ} = \frac{\sqrt{5} - 1}{4}, cos⁡36∘=5+14\cos 36^{\circ} = \frac{\sqrt{5} + 1}{4}, tan⁡22.5∘=2−1\tan 22.5^{\circ} = \sqrt{2} - 1. Allied angles. Every angle can be reduced to an acute one. Rule: for nπ2±θ\frac{n\pi}{2} \pm \theta, if nn is odd the ratio changes to its co-ratio (sin⁡↔cos⁡\sin \leftrightarrow \cos, tan⁡↔cot⁡\tan \leftrightarrow \cot, sec⁡↔csc⁡\sec \leftrightarrow \csc); if nn is even it stays the same. The sign is the sign of the original ratio in the quadrant where nπ2±θ\frac{n\pi}{2} \pm \theta lies (treating θ\theta as acute). Examples: sin⁡(π−θ)=sin⁡θ\sin(\pi - \theta) = \sin\theta; cos⁡(π+θ)=−cos⁡θ\cos(\pi + \theta) = -\cos\theta; tan⁡(π2+θ)=−cot⁡θ\tan(\frac{\pi}{2} + \theta) = -\cot\theta; cos⁡(3π2−θ)=−sin⁡θ\cos(\frac{3\pi}{2} - \theta) = -\sin\theta. Also sin⁡(−θ)=−sin⁡θ\sin(-\theta) = -\sin\theta and cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos\theta (sine odd, cosine even). Key Takeaways: - π\pi rad =180∘= 180^{\circ}; s=rθs = r\theta; sector area 12r2θ\frac{1}{2}r^{2}\theta. - Point on unit circle (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta) defines all ratios for any θ\theta. - sin⁡2+cos⁡2=1\sin^{2} + \cos^{2} = 1; 1+tan⁡2=sec⁡21 + \tan^{2} = \sec^{2}; 1+cot⁡2=csc⁡21 + \cot^{2} = \csc^{2}. - ASTC signs; allied angles: odd multiple of π2\frac{\pi}{2} switches to the co-ratio. - sin⁡18∘=5−14\sin 18^{\circ} = \frac{\sqrt{5} - 1}{4}; tan⁡15∘=2−3\tan 15^{\circ} = 2 - \sqrt{3}. - sec⁡θ+tan⁡θ=1sec⁡θ−tan⁡θ\sec\theta + \tan\theta = \frac{1}{\sec\theta - \tan\theta}. - Pitfall: using degrees in arc length or calculus formulas. Brain-Teaser 1: If sec⁡θ+tan⁡θ=3\sec\theta + \tan\theta = 3, find sin⁡θ\sin\theta and the quadrant of θ\theta. Solution: sec⁡θ−tan⁡θ=13\sec\theta - \tan\theta = \frac{1}{3}. Adding and subtracting: sec⁡θ=53\sec\theta = \frac{5}{3}, tan⁡θ=43\tan\theta = \frac{4}{3}. So cos⁡θ=35>0\cos\theta = \frac{3}{5} > 0 and sin⁡θ=tan⁡θcos⁡θ=45>0\sin\theta = \tan\theta\cos\theta = \frac{4}{5} > 0: θ\theta is in the first quadrant. The reciprocal identity for sec⁡±tan⁡\sec \pm \tan avoids squaring and the extraneous roots that squaring would bring. Brain-Teaser 2: Evaluate sin⁡25∘+sin⁡210∘+⋯+sin⁡285∘+sin⁡290∘\sin^{2}5^{\circ} + \sin^{2}10^{\circ} + \cdots + \sin^{2}85^{\circ} + \sin^{2}90^{\circ}. Solution: Pair sin⁡2θ\sin^{2}\theta with sin⁡2(90∘−θ)=cos⁡2θ\sin^{2}(90^{\circ} - \theta) = \cos^{2}\theta: each pair sums to 11. The angles 5∘,…,85∘5^{\circ}, \ldots, 85^{\circ} are 1717 terms; they form 88 pairs (55 with 8585, ..., 4040 with 5050) plus sin⁡245∘=12\sin^{2}45^{\circ} = \frac{1}{2}. Adding sin⁡290∘=1\sin^{2}90^{\circ} = 1: total 8+12+1=9.58 + \frac{1}{2} + 1 = 9.5. Complementary pairing is the standard trick for such sums. Brain-Teaser 3: A wheel of radius 3535 cm turns through 150∘150^{\circ}. How far does a point on its rim travel, and what area does the spoke sweep? Solution: 150∘=5π6150^{\circ} = \frac{5\pi}{6} rad. Arc length =35×5π6=175π6≈91.6= 35 \times \frac{5\pi}{6} = \frac{175\pi}{6} \approx 91.6 cm. Sector area =12×352×5π6=6125π12≈1603.5= \frac{1}{2} \times 35^{2} \times \frac{5\pi}{6} = \frac{6125\pi}{12} \approx 1603.5 cm2^{2}. Converting to radians first is what makes the simple formulas valid.

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Trigonometric Ratios: Radian Measure, the Unit Circle, Signs and Allied Angles — DPP

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