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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?

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.2: Trigonometric Equations
Problem 104E
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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?
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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