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e) triangle BOA is isosceles with altitude OM f) triangle COA is similar to triangle QOP g) demonstration h) lim h-->0 &

Posted: Thu Jun 02, 2022 12:23 pm
by answerhappygod
e) triangle BOA is isosceles with altitude OM
f) triangle COA is similar to triangle QOP
g) demonstration
h) lim h-->0 🔼tan t/🔼t = lim h-->0 (cos(h/2)•sec t•AB)/(cos(h+t)•h) = sec^2 t
E Triangle Boa Is Isosceles With Altitude Om F Triangle Coa Is Similar To Triangle Qop G Demonstration H Lim H 0 1
E Triangle Boa Is Isosceles With Altitude Om F Triangle Coa Is Similar To Triangle Qop G Demonstration H Lim H 0 1 (66.52 KiB) Viewed 45 times
E Triangle Boa Is Isosceles With Altitude Om F Triangle Coa Is Similar To Triangle Qop G Demonstration H Lim H 0 2
E Triangle Boa Is Isosceles With Altitude Om F Triangle Coa Is Similar To Triangle Qop G Demonstration H Lim H 0 2 (66.52 KiB) Viewed 45 times
272. Let A = (cost, sin t) represent an object moving counterclockwise at 1 unit per second on the unit circle. During a short time interval h, the object moves from A to B. To build a segment that represents A tant, a new type of projection is needed (keep in mind that sin was analyzed using a projection onto the y-axis in #254). Thinking of it as a wall, draw the line that is tangent to the unit circle at D = (1,0), then project A and B radially onto P and Q, respectively. In other words, OAP and OBQ are straight. As t increases from 0 to, P moves up the wall. Mark C on segment BQ so that segments AC and PQ are parallel. Also, mark the midpoint, M, of the segment AB. Explain why (a)-(f) are true. B M (e) AB = 2AM = 2 sin (h/2) (f) PQ = AC sect Working with AABC, apply the Law of Sines along with other trigonometric identities in order to conclude that: (g) AC = cos (h/2) cos (t + h) AB. (h) Now evaluate lim A tant, with At = h, to find the derivative of the tangent function. h→0 (i) What does your answer say about the derivative of y = tan x? Compare your answer with the graph drawn in #256.