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

Use the following trivial identity:

cos (0) = 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

= 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