What is the value of ln(1)*(1)^0+e^0*2*sin(0)+cos(0)*2*e^0 ?

The value of ln(1)*(1)^0+e^0*2*sin(0)+cos(0)*2*e^0 is 2

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ln(1)(1)^0+e^02*sin(0)+cos(0)2e^0

ln (1) \cdot (1)^{0} + e^{0} \cdot 2 \cdot sin (0) + cos (0) \cdot 2 \cdot e^{0}

2

Solution steps

ln (1) (1)^{0} + e^{0} \cdot 2 sin (0) + cos (0) \cdot 2 e^{0}

= ln (1) \cdot 1^{0} + e^{0} \cdot 2 sin (0) + cos (0) \cdot 2 e^{0}

Use the following trivial identity:

sin (0) = 0

sin (0)

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

Use the following trivial identity:

cos (0) = 1

cos (0)

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 1

= ln (1) \cdot 1^{0} + e^{0} \cdot 2 \cdot 0 + 1 \cdot 2 e^{0}

Simplify

ln (1) \cdot 1^{0} + e^{0} \cdot 2 \cdot 0 + 1 \cdot 2 e^{0} : 2

ln (1) \cdot 1^{0} + e^{0} \cdot 2 \cdot 0 + 1 \cdot 2 e^{0}

Apply rule

1^{a} = 1

1^{0} = 1

= 1 \cdot ln (1) + 2 \cdot 0 \cdot e^{0} + 1 \cdot 2 e^{0}

Apply rule

a^{0} = 1, a \neq = 0

e^{0} = 1

= 1 \cdot ln (1) + 1 \cdot 2 \cdot 0 + 1 \cdot 2 \cdot 1

ln (1) \cdot 1 = 0

ln (1) \cdot 1

Simplify

ln (1) : 0

ln (1)

Apply log rule:

lo g_{a} (1) = 0

= 0

= 0 \cdot 1

Apply rule

0 \cdot a = 0

= 0

= 0 + 1 \cdot 2 \cdot 0 + 1 \cdot 2 \cdot 1

Simplify

0 + 1 \cdot 2 \cdot 0 + 1 \cdot 2 \cdot 1

1 \cdot 2 \cdot 0 = 0

1 \cdot 2 \cdot 0

Apply rule

0 \cdot a = 0

= 0

1 \cdot 2 \cdot 1 = 2

1 \cdot 2 \cdot 1

Multiply the numbers:

1 \cdot 2 \cdot 1 = 2

= 2

= 0 + 0 + 2

Add the numbers:

0 + 0 + 2 = 2

= 2

= 2

= 2

2 cos^{2} (3 7^{\circ}) - 1

- e^{c o s (π)} - (- e^{c o s (0)})

1 + tan^{2} (31 5^{\circ})

- cos (4 \frac{π}{2})

arctan (\frac{16}{32})

What is the value of ln(1)*(1)^0+e^0*2*sin(0)+cos(0)*2*e^0 ?
The value of ln(1)*(1)^0+e^0*2*sin(0)+cos(0)*2*e^0 is 2