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Findℒ{f(t)}by first using a trigonometric identity. (Write your answer as a function of s.)f(t) = 12 cost −π6

User Stackich
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1 Answer

5 votes

Answer:


L(f(t)) = (6)/(S^2+1) [√(3) \ S +1 ]

Explanation:

Given that:


f(t) = 12 cos (t- (\pi)/(6))

recall that:

cos (A-B) = cos AcosB + sin A sin B


f(t) = 12 [cos\ t \ cos (\pi)/(6)+ sin \ t \ sin (\pi)/(6)]


f(t) = 12 [cos \ t \ (3)/(2)+ sin \ t \ sin (1)/(2)]


f(t) = 6 √(3) \ cos \ (t) + 6 \ sin \ (t)


L(f(t)) = L ( 6 √(3) \ cos \ (t) + 6 \ sin \ (t) ]


L(f(t)) = 6 √(3) \ L [cos \ (t) ] + 6\ L [ sin \ (t) ]


L(f(t)) = 6 √(3) (S)/(S^2 + 1^2)+ 6 (1)/(S^2 +1^2)


L(f(t)) = (6 √(3) +6 )/(S^2+1)


L(f(t)) = (6( √(3) \ S +1 )/(S^2+1)


L(f(t)) = (6)/(S^2+1) [√(3) \ S +1 ]

User Isatsara
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