Why is cos(0) = 1?

Name: Why is cos(0) = 1?
Uploaded: 2020-05-01T12:45:59+00:00
Channel: Joey Tang
Description: Why is cos(0) = 1?

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Keith LeannonLv2

24 May 2019

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Joey Tang

Bachelor’s Degree in Chemistry from McMaster University

1 May 2020

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11. Let f(1) = cos 1. Use the definition of the derivative to prove that f'(:1) = -sin . Solution: We have · f'(:1) = lim cos(2+h) - cos(2) +0 cos(1) cos(h) - sin() sin(h) - cos(1) = lim h0 sin / – Jin contar) (cos(1) = -1) = sin(e) (sinm) h0 sin(h) cos(h) - = lim cos(2) h cos(h) - 1 = cos(x) lim h- oh = cos(x)0 - sin(x)(1) = - sin h c) lim 0 / h

15. 1 - cos x - x sinx lim 0+0+ sina x + 2 cos x - 2 @ -o NIN

[1] 1. (a) Solve 2 – 31 < 7. Solution: We have – 7<-3 < 7 which gives -4 0 on (-0, 1] U [2,). Next, we need to see where Vr2 – 3r +2 -1 > 0. If r2 - 3.c + 2 – 1 > 0, then V.22 – 3.0 +2 > . Observe that this is definitely true if r < 0. If r > 0, then squar- ing both sides we get 22 - 3.+2 > 1 2 > 3. V A VICO Thus, the domain of f is (-00, ). [2] (c) Find the exact value of tan(sec-14). Solution: Let y = sec-14. Then, sec y = 4. From the diagram we get that tan(sec-? 4) = tan y = 15 [2] (d) Find all r such that sin(2.x) = COS I. Solution: We have 2 sin r cos r = COS I, SO 0 = 2 sin Cos I -Cos I = cos .r (2 sin c - 1). Hence, sin(2x) = cost when cos x = 0, or when sin r = Thus, r= +ik, kez, I= +2nk, ke Z, or r= +2nk, kez [2] (e) State the precise mathematical definition of lim f(x) = 0. Solution: For every e > 0 there exists a 8 >0 such that if 0 < x-al <, then f(c) >

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