HW_1_2_Solutions

# HW_1_2_Solutions - CE 221 Homework 1 — Due Wednesday 1...

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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—60-35=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 LABc-zzeco [BARAACF 1 Emma; + 1 600 = {Zr 1 BCMLBAc + LABL = Igo‘ Ma {um of *AL I‘M-WW 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 2-5 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: dtB-202—~2-6 2+4-6 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 20.51-Y=2 4—1—12 7 Hence, 111“3 0,200.23 2 =20.51-2=2.4—20.4-2=6 4 —1—1 7 —1.6 2 —0.6 7 —5 ...
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HW_1_2_Solutions - CE 221 Homework 1 — Due Wednesday 1...

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