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Unformatted text preview: Math 117 Assignment #3 - Fall 2008 due October 2nd or 3rd - A/ Suggested Problems: Adams §P.7 # 1—30, §3.5 # 1—16, §3.6 #7ab. B / For Submission: 1. Adams, §P.7 #13,15.
2. Adams, §3.5 # 13,15,17 (see Example 3, p.189) 3. Rewrite the following expressions in the form of a single sine function (of the form
r sin(l9 — (1)): a) sin 6 + 8 cos 0
b) 3 cos (9 — sin 0. 4. You may have been introduced to the hyperbolic functions in class. These are
deﬁned as the even and odd components of the natural exponential function: ex + 6"“5 ex — e“
coshz= ————-, sinhx= ——
(You can see that cosh z is even, sinhz is odd, and cosh cc + sinh a: = ex.)
From these deﬁnitions, show that a) sinh(:r + y) = sinh :1: cosh y + cosh z sinh y
sinh(3$) = 3 sinh z + 4 sinh3 2:.
In fact, the hyperbolic functions obey a set of identities very similar to that of the trigonometric functions. To obtain one from the other, we-simply change the
sign of every product of two Sines; this is “Osborn’s Rule”. cos2 x + sin? cc = 1 ——> cosh2 a: — sinh2 :1: = 1 1 + tan2 z = 3602.73 ——> 1 — tanh2 z = sech2$ cos(z + y) = cos ac cosy —— sin a: sin y ——> cosh(a: + y) = cosh a; cosh y + sinh zsinh y
cos2z= -21;(1+cos2z) ——>cosh2z= %(1+cosh2m) etc. Math 117 — Assignment #3 Page 2
5. Consider the function f (:13) = 3:3 + x. This is a one-to—one function (this should
be obvious, if you think of what the graph must look like), so we know it must
have an inverse, but it isn’t obvious that there’s a convenient way to write it.
Surprisingly, the hyperbolic sine function can help, because of one of the identities above. Follow these steps to ﬁnd f ‘1: L“ y = 173 + 9:. Then set z = ‘99:, and show that 4z3 + 3z = #9 Now let 2 = sinh 6, and apply the identiy you proved in #4b above. You can then
solve for 6, and reverse the changes of variables to ﬁnd :1: as a function of y. ...
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