Question

L {cos 6t cos 4t sin t}​

Answer 1

(sin

2

t+cos

2

t)

3

−3sin

2

tcos

2

t(sin

2

t+cos

2

t)−1

(sin

2

t+cos

2

t)

2

−2sin

2

tcos

2

t−1

since (a+b)

2

=a

2

+2ab+b

2

and (a+b)

3

=a

3

+3a

2

b+3ab(a+b)

=

−3sin

2

tcos

2

t

−2sin

2

tcos

2

t

=

3

2

verified_toppr

Answer verified by Toppr

View

233 Views

Was