What is 1/81 (sin(9(t-x)))+9xcos((t-x)9) ?

The solution to 1/81 (sin(9(t-x)))+9xcos((t-x)9) is 1/81 sin(9(t-x))+9xcos((t-x)*9)

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simplify 1/81 (sin(9(t-x)))+9xcos((t-x)9)

simplify

\frac{1}{81} (sin (9 (t - x))) + 9 x cos ((t - x) 9)

\frac{1}{81} sin (9 (t - x)) + 9 x cos ((t - x) \cdot 9)

Solution steps

\frac{1}{81} (sin (9 (t - x))) + 9 x cos ((t - x) \cdot 9)

Apply rule:

(a) = a

(sin (9 (t - x))) = sin (9 (t - x))

= \frac{1}{81} sin (9 (t - x)) + 9 x cos ((t - x) \cdot 9)

- \frac{n ( d ^{2} v + e d v + d v f - d b + e v f - e b )}{m c ( d + e )}

(1.6707100000000 \dots + 0.8976700000000 \dots \cdot i)^{4}

\frac{3 s ^{2} + e ^{- 4 s}}{s ( s ^{2} + 1 )}

(4 - 3 i) + (- 1, 5) - (8, - 4)

(A + (C \cdot cos (y)) + (D \cdot cos (z)) - (B \cdot cos (x)))^{2}

What is 1/81 (sin(9(t-x)))+9xcos((t-x)9) ?
The solution to 1/81 (sin(9(t-x)))+9xcos((t-x)9) is 1/81 sin(9(t-x))+9xcos((t-x)*9)