(a) Prove the Calculus identity sin- (a) + sin-(VT- 2²) = 5, z€ (0,1], where sin-, denotes the usual real inverse trig functions.
(a) Prove the Calculus identity sin- (a) + sin-(VT- 2²) = 5, z€ (0,1], where sin-, denotes the usual real inverse trig functions.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![(a) Prove the Caleulus identity
sin-(x) + sin-(V1– 2²) = 5
r€ (0, 1),
%3D
where sin-, denotes the usual real inverse trig functions.
(b) Create and submit a plot of sin (r)+ sin-(VT- r²). Explain why the identity
is false if a < 0.
(c) In this question, all integrals are Calculus-type real integrals. Let
f(t) = VT-t,-1sts1,
and consider the substitution s = f(t). Taking a derivative gives
tdt
ds = -
Hence,
dt
ds
By the substitution rule for definite integrals,
dt
ds
dt
dt
Vī
dt
dt
dt
%3D
VI- 12
0.
However, as in (b), the identity is false if r < 0. Explain what went wrong.
(d) We now move off the real line and into the complex plane. Suppose wi + w = 1
and that wi, w2 e RHP (open right half-plane). Let C1, C2 be curves that,
respectively, connect 0 to w, and w, and that avoid K = (-00,-1]U[1, 00). Use
the substitution rule for complex line integrals to show that, with appropriately
chose branches of (1 – 22)/2, we have
%3D
dç
Jo, (1- (?)1/2 + Jc, 1 – (*)!/2 = 7
(1- C2)1/2 = 2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb56d042f-326e-4d12-a439-025aa89b72fb%2Fc7b8f1b5-f791-4557-8649-68047e93a0a5%2F6caw2m_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(a) Prove the Caleulus identity
sin-(x) + sin-(V1– 2²) = 5
r€ (0, 1),
%3D
where sin-, denotes the usual real inverse trig functions.
(b) Create and submit a plot of sin (r)+ sin-(VT- r²). Explain why the identity
is false if a < 0.
(c) In this question, all integrals are Calculus-type real integrals. Let
f(t) = VT-t,-1sts1,
and consider the substitution s = f(t). Taking a derivative gives
tdt
ds = -
Hence,
dt
ds
By the substitution rule for definite integrals,
dt
ds
dt
dt
Vī
dt
dt
dt
%3D
VI- 12
0.
However, as in (b), the identity is false if r < 0. Explain what went wrong.
(d) We now move off the real line and into the complex plane. Suppose wi + w = 1
and that wi, w2 e RHP (open right half-plane). Let C1, C2 be curves that,
respectively, connect 0 to w, and w, and that avoid K = (-00,-1]U[1, 00). Use
the substitution rule for complex line integrals to show that, with appropriately
chose branches of (1 – 22)/2, we have
%3D
dç
Jo, (1- (?)1/2 + Jc, 1 – (*)!/2 = 7
(1- C2)1/2 = 2
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