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Unformatted text preview: CE 221 Homework 1 — Due Wednesday January 21, 2009 1. Proof the following trigonometric identity:
2
Tan (201) = tan :1
1— tan a
8i ( )
sin(2a) 2sin(a)cos(a) 2 n VCW) 2tan(a)
tan(2a) : 2 2‘4— : I 2 = 2
cos(2a) cos (a)—sm (a) 1_sm (a/ l—tan (a)
cosz(a) 2. Proof that sin(a+b) = [sin(a)][cos(b)] + [cos(a)][sin(b)]
Consider the ﬁgure below. Let E =1 then, sin(a+b)=ﬁ=—B—D+ﬁ. IA '7 / O G Since 20CD = (1, ABCD = 90° —a
Here, E = EsiszCD) = Esmm" — a) = Ecosm) = sin(b)cos(a)
D_F = E 2 Emma) = cos(b)sin(a) Therefore, sin(a + b) = E = E + D_F = sin(a)cos(b) + cos(a) sin(b) 3. Proofs that for any triangle, the following ratios are equal (the sin
law).
AB/Sin(c) = AC/Sin(b) = BC/Sin(a) A From point A draw AA’ at 900 to BC. AA’ = AC(sin c) = AB(sin
b)
Arranging terms yields: AB/Sin(c) = AC/Sin(b) Repeat the process by drawing from point B a line at 900 to AC 4. Proof the following trigonometric identity: 1+ Sin(2a) Tan2(45 + a) =
l— Sin(2a) Since cos(2x) = l — 2sin2 (x) = Zoos2 (x) —l l— cos(2x) 1+ cos(2x) d 2 =
an cos (x) 2 sin2 (x) = Also recall that cos(90° + x) : —sin(x) sin2 (45° + a) _ l— cos(90° + 2a) _ 1+ sin(2a)
cos2 (45° + a) 1+ cos(90° + 2a) 1— sin(2cz) tan2(45° + a) = 5. Find the magnitude of the resultant force F of the given 45 and 55 pound forces. B =180—6035=850
F2 = 452 +552 —2*45*55*cos(85°)= 4618.6 F: 67.96 #= 68 #
6. Proof that sinza + cosza = 1.0
By definition: Sin a = Opposite/Hypothenus and Cos a = adjacent/hypothenus Squaring and adding yields: sinza + cosza = (opposite2 + adjacent2)/hypothenus2 By Pythagoras theory (0 osite2 + ad'acentz) : hy othenus2
PP J P  2 2
Hence: $11] a + cos a = 1.0 7. Proof that the sum of the interior angles of a triangle is 180°. A LABczzeco [BARAACF 1 Emma; + 1 600 = {Zr 1 BCMLBAc + LABL = Igo‘
Ma {um of *AL I‘MWW My!“ .f a {Amyé :lgp. I
r l 8. Proof that the sum of the interior angles of a hexagon (6 sided) is 720°. D From an interior point (such as G) draw 6 triangles. The sum of the interior angle of each triangle = 180°. For the six triangles, the sum of the interior angles is 6(180) = 10800. Subtract the sum of the interior angles at G of 3600 yields: 1080 — 360 2 720°. Proof that cos(a + b) = [cos(a)][cos(b)] — [sin(a)][sin(b)]
Consider the ﬁgure below. Let E :1 then, cos(a+b)=ﬁ=E—F—G. Since AOCD = a , ABCD = 90° —a Here, E = C_D = EcosMBCD) = Ecos(90° — a) = EEsinm) = sin(b)sin(a)
E = Room) = cos(b)cos(a) Therefore, cos(a + b) = E = E — E = cos(a) cos(b) — sin(a) sin(b) 10. Proof that for any triangle, the following relationship is true BC2 = BA2 + AC2 — 2(BA)(AC)(cos(a)) Got: Eo‘wo‘ = (AB'Mf‘L C91
86“: m mﬂ c 05 MW”
5M. Ao‘: Ac“~ CD‘ (Rf/1W) 95": ABE} Pm" co”? CD:k " WWW) ; AB? AC" MEMO
Av; Acmsﬂ ~_ M“
Brake Mfg“ LMHM? BC" W‘Mcz ~2LRB)LAc)fw1Ca)]
7M: ’ 11. Obtain the determinants of the following matrices. [A]=[j3 Z] Sol: det[A] = (1)  (6) — (2)  (—3) = 12 123
BL:3 3 2
2—51 SI dt[B] (1) 3 (2) 3 2+(3) 3 3 48
o: =   ' ' =—
e —51 2 1 25
1 1 0 l
2 —5 l
C =
[ ] 0 2 4 7
0 —l 6 8
S01:
3 —5 1 2 —5 1 2 3 1 2 3 —5
det[C]=(l)2 4 7—(1)~0 4 7+(0)0 2 7—(l)0 2 4:89
—1 6 8 0 6 8 0 —l 8 0 —1 6 12. Obtain [A]T for the given [A] matrix. 4 3 —2
[A]: —2 3 4
2 —6 5 4 —2 2 sm;MY= 3 3 —6 13. Obtain the inverse [B]1 of the given [B] matrix. 2—24
[B]—[6 0 2
2 0 4 Sol: dtB202—~26 2+46 0—40 Ci]—()'04()24()20
02 62 60 T
i i WI i l I
04 24 20 d,[B]_ _1—24 24 _12 —2 a] ‘()'o 4 24 ()2 0
—24 24 2—2
I I “Di l i i
0 2 62 6 0 det[B] _[ 1 0 0.2 —0.1
[B]1 = 3ij = —0.5 0 0.5
0 —0.1 0.3 14. Obtain matrix [C] by multiplying matrices [A] and [B]. 2—3 [c]=[41*[31=[f ‘05 j} 2 4
6 5 [C] 2 [(2] (2)+ ( 5) (2)+ (6) (6) (2) (— 3)+ ( 5) (4)+ (6) (5)]
(1) (2)+ (0) (2) + (4) (6) (1) (— 3) + (0) (4) + (4) (5) _304
_2617 15. Solve the following equations using matrix operation. Verify your answer using the traditional solution of three simultaneous equaﬁons
X+Y+Z=3
2X+05Y+Z=2
4X—Y—Z=7
Sol: Writing the above equations in matrix form results in... 111 X 3 20.51Y=2 4—1—12 7 Hence, 111“3 0,200.23 2
=20.512=2.4—20.42=6
4 —1—1 7 —1.6 2 —0.6 7 —5 ...
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 Summer '08
 Buch
 Addition, Pythagorean Theorem, Following, Interior, Summation

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