A geometric proof that the derivative of sin x is cos x. At the start of the lecture we saw an algebraic proof that the derivative of sin x is cos x. While this proof was perfectly valid, it was somewhat abstract – it did not make use of the deﬁnition of the sine function. sin θ The proof that lim = 1 did use the unit circle deﬁnition of the sine of θ →0 θ an angle. It also showed that when x = 0 the derivative of sin x is 1: d sin x|x=0 dx = = = = sin(0 + Δx) − sin 0 Δx sin Δx − 0 lim Δx→0 Δx sin Δx lim Δx→0 Δx 1. lim Δx→0 We’ll now prove that the derivative of sin θ is cos θ directly from the deﬁnition |opposite| of the sine function as the ratio of the side lengths of a right |hypotenuse| triangle. Q P Δθ θ 1 O Figure 1: Point P has vertical position sin θ. We start with a point P on the unit circle centered at O and the angle θ associated with P . As indicated in Figure 1, sin θ is the vertical distance between P and the x-axis. Next, we add a small amount Δθ to angle θ; let Q be the point on the unit circle at angle θ + Δθ. The y-coordinate of Q is sin(θ + Δθ). To ﬁnd the rate of change of sin θ with respect to θ we just need to ﬁnd the rate of change of y = sin θ. 1 Q R Δy P � Figure 2: When Δθ is small, P Q≈ P Q. Find dy dθ . As shown in Figure 2, Δy = |P R| and segment P Q is a straight line approx imation of the circular arc P Q. If Δθ is small enough, segment P Q and arc P Q are practically the same, so |P Q| ≈ Δθ. We’re trying to ﬁnd Δy. Since we know the length of the hypotenuse P Q, all we need is the measure of � QP R to solve for Δy = |P R|. Since Δθ is small, segment P Q is (nearly) tangent to the circle, and so angle � OP Q is (nearly) a right angle. We know that P R is vertical, we know that θ is the angle OP makes with the horizontal, and we can combine these facts to prove that � RP Q and θ are (nearly) congruent angles.1 The arc length Δθ is approximately equal to the length |P R| of the hy potenuse and angle RP Q is approximately equal to θ. By the deﬁnition of the R| cosine function we get cos θ ≈ |PΔθ . But |P R| is just the vertical distance be tween Q and P , which is just the diﬀerence between sin(θ + Δθ) and sin θ. In other words, when Δθ is very small, cos θ ≈ sin(θ + Δθ) − sin θ . Δθ As Δθ approaches 0, segment QP gets closer and closer to arc QP and angle θ) QP O gets closer and closer to a right angle, so the value of (sin(θ+Δθ)−sin gets Δθ closer and closer to cos θ. We conclude that: sin(θ + Δθ) − sin θ = cos θ Δθ →0 Δθ lim and thus that the derivative of sin θ is cos θ. 1 Professor Jerison does this by rotating and translating angle θ to coincide with angle RP Q. Another way to see this is to extend segment RP until it intersects the horizontal line through O at point S, then note that m� RP Q + m� QP O + m� OP S = π and also ∼ m� P SO, we get m� RP Q = ∼ θ. (If θ > π/2 a θ + m� P SO + m� OP S = π. Since m� QP O = diﬀerent, but similar, argument applies.) 2