we can write the given function as: x(t) = 3 + 4 cos (20nt + ) 2 sin (60nt -)-4 cos (70nt – π) As we know that, cos x = SO, 1-3 +48 (2011-2) + 6 -1(20+2), 20 (60m - 2) + ²/(60m-2) (20nt+)+(20nt (60nt-7)+7(60nt-) +e +e ²1-4 x(t)=3+4 2 2 π x(t)=3+2 +e −3+2[6 (20m +₹) + ¯/(20m + H)] + [✔✅(60m-) + 7/(60nt-)]_2[@(70nt-n) + @~7/(70nt-n)] -¹]+[60nt./+160mt. ²]-2[ei/70 nt . e¯jin+e¬/70nt, gin] x(t)=3+ −3+2/20mt. +²/20nt. .e 20/70 m.e-/+2e-/70mt.e ·ejn x(t)=3+2e/20mt.e +2e-/20mt.e +e/60 mt. e te-jx 2 -1/2 te +e-j60 mt.e (70nt-n) + e-(70nt-n) 2 Isn't it supposed that the value 2sin(60nt) turns into 2cos(60nt) after subtracting π/2 and that the value of -4cos(70nt) turns into a positive after adding to it?
we can write the given function as: x(t) = 3 + 4 cos (20nt + ) 2 sin (60nt -)-4 cos (70nt – π) As we know that, cos x = SO, 1-3 +48 (2011-2) + 6 -1(20+2), 20 (60m - 2) + ²/(60m-2) (20nt+)+(20nt (60nt-7)+7(60nt-) +e +e ²1-4 x(t)=3+4 2 2 π x(t)=3+2 +e −3+2[6 (20m +₹) + ¯/(20m + H)] + [✔✅(60m-) + 7/(60nt-)]_2[@(70nt-n) + @~7/(70nt-n)] -¹]+[60nt./+160mt. ²]-2[ei/70 nt . e¯jin+e¬/70nt, gin] x(t)=3+ −3+2/20mt. +²/20nt. .e 20/70 m.e-/+2e-/70mt.e ·ejn x(t)=3+2e/20mt.e +2e-/20mt.e +e/60 mt. e te-jx 2 -1/2 te +e-j60 mt.e (70nt-n) + e-(70nt-n) 2 Isn't it supposed that the value 2sin(60nt) turns into 2cos(60nt) after subtracting π/2 and that the value of -4cos(70nt) turns into a positive after adding to it?
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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i need the answer quickly
![x(t) = 3+4 cos(20nt +30°) + 2 sin(60rt) - 4 cos (70ft)
we can write the given function as :
|x (t) = 3 + 4 cos (20πt + 7) — 2 sin (60πt —
2 sin (60nt -)-4 cos (70πt — ñ)
exte-jx
As we know that, cos x = 2
So,
[ (20m + 3) + 6071(20m+3)], [6 (60m-3) + ²/(60m - 2)
+e
+e
2
N
x(t)=3+4
|x (t) = 3 + 2[/(20 mt+) + ¯i (20 mt +)]+[(60nt-)+(60nt-)-2[/(70nt-n) + −/(70nt-n)]
+e
→/]+[/60nt. + ¯/60mt. ²]2[/70 nt .e-j+e-170mt, m]
+6²
/20nt,
-/20nt.6
x(t)=3+2e2
x(t)=3+2/20nt,
-i te/60mt.e
2²-20/70 m.e-j+2e-/70m.in
+2e-/20πt.e
T
mt.e
-1/2 +6²
21-4
-j60 nt.
e¡ (70nt − n) +ẹ¯¡ (70nt − n)
2
Isn't it supposed that the value 2sin(60nt)
turns into 2cos(60nt) after subtracting π/2
and that the value of -4cos(70nt) turns into
a positive after adding to it?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F013331d1-1dfa-497a-847a-05c3002de4bb%2F79d7ff39-2312-44bb-a4b6-36b0833cb5b7%2Fdc0e8cc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:x(t) = 3+4 cos(20nt +30°) + 2 sin(60rt) - 4 cos (70ft)
we can write the given function as :
|x (t) = 3 + 4 cos (20πt + 7) — 2 sin (60πt —
2 sin (60nt -)-4 cos (70πt — ñ)
exte-jx
As we know that, cos x = 2
So,
[ (20m + 3) + 6071(20m+3)], [6 (60m-3) + ²/(60m - 2)
+e
+e
2
N
x(t)=3+4
|x (t) = 3 + 2[/(20 mt+) + ¯i (20 mt +)]+[(60nt-)+(60nt-)-2[/(70nt-n) + −/(70nt-n)]
+e
→/]+[/60nt. + ¯/60mt. ²]2[/70 nt .e-j+e-170mt, m]
+6²
/20nt,
-/20nt.6
x(t)=3+2e2
x(t)=3+2/20nt,
-i te/60mt.e
2²-20/70 m.e-j+2e-/70m.in
+2e-/20πt.e
T
mt.e
-1/2 +6²
21-4
-j60 nt.
e¡ (70nt − n) +ẹ¯¡ (70nt − n)
2
Isn't it supposed that the value 2sin(60nt)
turns into 2cos(60nt) after subtracting π/2
and that the value of -4cos(70nt) turns into
a positive after adding to it?
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