What is (1/2+3j)(cos(2t)-jsin(2t)) ?

The solution to (1/2+3j)(cos(2t)-jsin(2t)) is ((6j+1)(cos(2t)-jsin(2t)))/2

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simplify (1/2+3j)(cos(2t)-jsin(2t))

simplify

(\frac{1}{2} + 3 j) (cos (2 t) - j sin (2 t))

\frac{( 6 j + 1 ) ( cos ( 2 t ) - j sin ( 2 t ) )}{2}

Solution steps

(\frac{1}{2} + 3 j) (cos (2 t) - j sin (2 t))

Join

\frac{1}{2} + 3 j : \frac{6 j + 1}{2}

= \frac{6 j + 1}{2} (cos (2 t) - j sin (2 t))

Apply the fraction rule:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( 6 j + 1 ) ( cos ( 2 t ) - j sin ( 2 t ) )}{2}

\frac{4.4}{l ( 1.1 ) + 0.56}

\frac{50000}{s ( s + 10 ) ( s + 50 )} \cdot (\frac{0.0899 ( s + 9.81 )}{s + 44.102})

\frac{10 ( π ^{3} + 25 ( π ^{2} + 6 ) )}{π ^{3}}

(e^{y} - x)^{2}

\frac{( 3 x - 1 ) ( x + 2 )}{( x - 3 ) ^{2}}

What is (1/2+3j)(cos(2t)-jsin(2t)) ?
The solution to (1/2+3j)(cos(2t)-jsin(2t)) is ((6j+1)(cos(2t)-jsin(2t)))/2