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Unformatted text preview: Definition ofthe Inverse Trig Functions: Inverse cosine denoted cos’lx = um’ ue number in 0,127 whose cosine is x l
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Inverse sine denoted sin x—  unique number 1n —— ,~—] whose sine is x T \V
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Inverse tangent denoted tan’lx = unique number 1n [—E 3] whose tangent is )1 Let’s look at our calculator and how you can use it to find the' 1nverse ofthe basic
trig functions. Ex: Find cos—If .4) by using your calculator. sew"
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Ex: Let sinx = E. Solve for all real x l i Note: If you are not given domain restriction, you have to solve for all x. Otherwise,
stick to given x boundary SlM ‘1‘"
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.2“ 1 —«F Ex: Find cos (——)
if“) 2 The Inverse Trig Functions (8.5) In this section we want to study the inverse trigonometric functions
y = sin“l(x), y = cos'1(x), and y = tan"(x) with more conceptual foundations. Recall: 1) In order for y = sinx and y : tanx to have an inverse, we have to restrict
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their domains to [3 3] and for y = cosx to have an inverse, we had to restrict the domain to [031:] By restricting the domain we force the trigonometric functions
y = sinx, y = cosx and y = 'tanx to become one to one. 2] We will able to obtain the graphs of the inverse functions by reﬂecting the graphs
of the basic trigonometric functions across the line y = x (Look below for graphical
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Q I ) 5112 “72’ Algebraically to find an inverse we will have to perform the following steps: 1) Let y = sinx
___.——'—7
___._———vo Switch the X and y variables
____________ X= siny
ﬁJW—ﬂ ﬂ; 2) Solve for the newry gal (x ) 2% (9/36) to get y by itself we have to take the" Inverse of both sides in step 1. «:Sm >( Therefore, )2 = sin—Xx) Domain: [~— 1 1 [1 Range: .. if" j
Y: tV7, ) ’7.“ Definition: sin(sin’ x) = x for all x in sin—1(sin x — x for all x i is
:‘i Similarly for y = cos—1x Dornain: E... i! U
Range: [0‘63 ' Definition: cosmos—Ix) = x for all x in [—1 ,1] cos—1(cosx) m x for all X in [0,71] Ex:Find£gs(co_;s"l(§))= E ugfwﬁ m Mmkm Q)OO\"€>
3 x .4
Find cos(sin“‘(—:~))= _ ‘ M 5m 2:) 5 E},
W ”6‘9? ‘ ﬁivx (911K ‘ (2:9 \.,—, 5M9)
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‘ t} w(ﬂs(1ﬁL ) «397$? BER) :: ‘ ”Va ' ' _ »—I =
E x: arcos[cos J '3 i x
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I For y: tan (x) find Domain. C. 0% (>0) Range. [MW T‘ i J . tan(tan x): "xforaalxa tan'1(tan x) = x for all x Ex: Findarc[tan0)= @ WK (W O 3 4: EX: ' l , '
Findtan'lﬂ):  “ﬂy/Ll
arctan(tan(i)) = —— [T/n . : 7 .. ’7 Evaluate the following without using your calculator: w ~—[
tan(sin'l(£)): _ , 0i} Mm (25L) 5 5m (3m‘6ﬂq) Does norMist 319L651?) "E
_ to  a " 2,
Evaluate without using your calculator ' 4:1": I
3 sin"l(sin(4?x)) ‘— Slh VHU 3., 9”};
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Note: Remember to pay extra attention to the domain of the' Inverse sine functngé SW.— l "I“: 5.2, text  ‘ '
2 _1 7: 9mg \ 7 . $605413“; 5.0"" _ 7‘ Graph: y—FFCOS (1 Aﬂb TVJ/ are"COS “‘(X*‘D yr OUCUCLS W}! ecjw‘o‘b j 4/
. ‘ . . W
. ")K ‘ ‘ L Chapter 9.1: The law of sines and cosines In this section we will discuss two laws, the law of sines and cosines that establishes
relationship between the sides and angels illMM (acute, right or obtuse
triangles). The purpose of these laws is to solve a triangle {find out the lengths of .7 R—h—E—‘m‘mmn
the thd the measurements of the three angels] when you know only some O‘ffh‘e side lengths or some of the angel measurements. You will have to
decide which of the laws is better to use giyen the provided information in the
question. The Law of sines: In any triangle, the ratio of the sine of an angle to the lengths of
the opposite side is constant. Low? 0.? SCI/be ‘.
94‘: You Wm Owe Angk—
W/ m CBWQS'FoucQIKﬁ s Aide . IPi‘oof: Raw,“ «0 AW.” 4% 0% "T‘l‘ a "'2. r f 2 3. {elmw
;D‘iude. Ml Melts 3% lialﬁ— “*9 $40ng : 9M0 .1 E‘Zé—B— 1’0“qu t“
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s. Note: To uSe law of Sines, you will need the measure of one angel and its. opposite
side plus another angel or side. EX: Suppose a=4cmy b=9cm and <B=600. Find the angels <A, <C, and side c 0 911.63 I: «163% Ex: Given < C = 45°, b = 4w/Eﬁ, c = 8ft. Solve the triangel. envy.“ Note: E it possible for this triangle to have two solutioggl Ex: Find the lengths a, b, c, and d in the fdlloVVing figure V The Law of cosines: In any triangle (acute, right or obtuse] the square of the length
of any side equals the sum of the squares of the lengths of the other minus twice the
product of the lengths of the other two sides times the cosine of their included
angels. @zb2+cz—2®os b2 = a2 + c2 —2accosB c2 = a2 + [:2 —2ab cosC
Note: 1) All three equations follow the same pattern
(square of the length of one side )2: [sum of the square of the other two sides) — [twice the product of the lengths of the other two sides times the cosine of their
included angel] 2) Pythagorian Theorem is a special case of the law of cosine.
For example, let <A = 900 Then the triangle becomes a right triangle Plugging into the law of cosine yields: a2 =b2+c2—2bc#s’9eo)=bé+cz at2 = b2 + c2 (The conclusion of the Pythagorian Theorem) Note: Law of cosine can be used when you know ASS or SSS of a triangle.  EX: <A = 550 , b = 9, (3:7. Find all other angels and sides. 9? \jou' Maul
€555 Qibic Use 3.00590
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3/43 Ex: Given a: 33, b = 7, and c = 37. Solve the triangle Remember: No ambiguous case for the law of cosine since cosx is positive in quadrant I and IV and quadrant IV angels are greater than 2700 which is not
possible for a triangel. ...
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 Spring '08
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