Answer: see proof below
<u>Step-by-step explanation:</u>
Use the Double Angle Identity: sin 2Ф = 2sinФ · cosФ
Use the Sum/Difference Identities:
sin(α + β) = sinα · cosβ + cosα · sinβ
cos(α - β) = cosα · cosβ + sinα · sinβ
Use the Unit circle to evaluate: sin45 = cos45 = √2/2
Use the Double Angle Identities: sin2Ф = 2sinФ · cosФ
Use the Pythagorean Identity: cos²Ф + sin²Ф = 1
<u />
<u>Proof LHS → RHS</u>
LHS: 2sin(45 + 2A) · cos(45 - 2A)
Sum/Difference: 2 (sin45·cos2A + cos45·sin2A) (cos45·cos2A + sin45·sin2A)
Unit Circle: 2[(√2/2)cos2A + (√2/2)sin2A][(√2/2)cos2A +(√2/2)·sin2A)]
Expand: 2[(1/2)cos²2A + cos2A·sin2A + (1/2)sin²2A]
Distribute: cos²2A + 2cos2A·sin2A + sin²2A
Pythagorean Identity: 1 + 2cos2A·sin2A
Double Angle: 1 + sin4A
LHS = RHS: 1 + sin4A = 1 + sin4A 
The equation y=mx+b
Slope= m
y-intercept= b
y= 2x+0 so y=2x
In an installment loan, a lender loans a borrower a principal amount P, on which the borrower will pay a yearly interest rate of i (as a fraction, e.g. a rate of 6% would correspond to i=0.06) for n years. The borrower pays a fixed amount M to the lender q times per year. At the end of the n years, the last payment by the borrower pays off the loan.
After k payments, the amount A still owed is
<span>A = P(1+[i/q])k - Mq([1+(i/q)]k-1)/i,
= (P-Mq/i)(1+[i/q])k + Mq/i.
</span>The amount of the fixed payment is determined by<span>M = Pi/[q(1-[1+(i/q)]-nq)].
</span>The amount of principal that can be paid off in n years is<span>P = M(1-[1+(i/q)]-nq)q/i.
</span>The number of years needed to pay off the loan isn = -log(1-[Pi/(Mq)])/(q log[1+(i/q)]).
The total amount paid by the borrower is Mnq, and the total amount of interest paid is<span>I = Mnq - P.</span>
At the end of the directions, it talks about parallel, perpendicular, or neither. When x=y, the lines are perpendicular and have infinite solutions. When they are parallel, there is no solution. You'll get an answer like 4=5. Neither means there is only one answer, and you will get one x and one y.
Answer:
Y E S : )
Step-by-step explanation:
