sin(x) =1 x→ 0 x lim In order to compute speciﬁc formulas for the derivatives of sin(x) and cos(x), we needed to understand the behavior of sin(x)/x near x = 0 (property B). In his lecture, Professor Jerison uses the deﬁnition of sin(θ) as the y-coordinate of a point on the unit circle to prove that limθ→0 (sin(θ)/θ) = 1. We switch from using x to using θ because we want to start thinking about the sine function as describing a ratio of sides in the triangle shown in Figure 1. The variable we’re interested in is an angle, not a horizontal position, so we discuss sin(θ)/θ rather than sin(x)/x. 1 θ sinθ arc length =θ Figure 1: A circle of radius 1 with an arc of angle θ. Our argument depends on the fact that when the radius of the circle shown in Figure 1 is 1, θ is the length of the highlighted arc. This is true when the angle θ is described in radians but NOT when it is measured in degrees. |opposite| Also, since the radius of the circle is 1, sin(θ) = equals the |hypotenuse| length of the edge indicated (the hypotenuse has length 1). In other words, sin(θ)/θ is the ratio of edge length to arc length. When θ = π/2 √rad, sin(θ) = 1 and sin(θ)/θ = 2/π ∼ = 2/3. When θ = π/4 rad, √ ∼ sin(θ) = 2/2 and sin(θ)/θ = 2 2/π = 9/10. What will happen to the value of sin(θ)/θ as the value of θ gets closer and closer to 0 radians? We see from Figure 2 that as θ shrinks, the length sin(θ) of the segment gets closer and closer to the length θ of the curved arc. We conclude that as θ → 0, 1 θ sin θ arc length =θ Figure 2: The sector in Fig. 1 as θ becomes very small sin θ → 1. In other words, θ sin(x) = 1. x→ 0 x This technique of comparing very short segments of curves to straight line segments is a powerful and important one in calculus; it is used several times in this lecture. lim