PLEASE HELP FAST!!! ILL MARK B.
Answer:
1 6/8
Step-by-step explanation:
add 1/8 to 5/8 by adding just the numerators
1 + 5 = 6
now you have 6/8. add the 1 and you have 1 6/8
hope this helped, have a good day :)
First list all the terms out.
e^ix = 1 + ix/1! + (ix)^2/2! + (ix)^3/3! ...
Then, we can expand them.
e^ix = 1 + ix/1! + i^2x^2/2! + i^3x^3/3!...
Then, we can use the rules of raising i to a power.
e^ix = 1 + ix - x^2/2! - ix^3/3!...
Then, we can sort all the real and imaginary terms.
e^ix = (1 - x^2/2!...) + i(x - x^3/3!...)
We can simplify this.
e^ix = cos x + i sin x
This is Euler's Formula.
What happens if we put in pi?
x = pi
e^i*pi = cos(pi) + i sin(pi)
cos(pi) = -1
i sin(pi) = 0
e^i*pi = -1 OR e^i*pi + 1 = 0
That is Euler's identity.
If tan (x)+pi/6=0
thn
tan (x)=-pi/6
take atan of both sides
x=atan (-pi/6)
x= -0.4823 rad
= -27.636 deg
Answer: see proof below
<u>Step-by-step explanation:</u>
Given: A + B + C = π → C = π - (A + B)
→ sin C = sin(π - (A + B)) cos C = sin(π - (A + B))
→ sin C = sin (A + B) cos C = - cos(A + B)
Use the following Sum to Product Identity:
sin A + sin B = 2 cos[(A + B)/2] · sin [(A - B)/2]
cos A + cos B = 2 cos[(A + B)/2] · cos [(A - B)/2]
Use the following Double Angle Identity:
sin 2A = 2 sin A · cos A
<u>Proof LHS → RHS</u>
LHS: (sin 2A + sin 2B) + sin 2C
![\text{Factor:}\qquad \qquad \qquad 2\sin C\cdot [\cos (A-B)+\cos (A+B)]](https://tex.z-dn.net/?f=%5Ctext%7BFactor%3A%7D%5Cqquad%20%5Cqquad%20%5Cqquad%202%5Csin%20C%5Ccdot%20%5B%5Ccos%20%28A-B%29%2B%5Ccos%20%28A%2BB%29%5D)
LHS = RHS: 4 cos A · cos B · sin C = 4 cos A · cos B · sin C 
