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KATRIN_1 [288]
3 years ago
6

A total of $12,000 is invested in two corporate bonds that pay 7.5% and 9% simple interest. The invest wants an annual interest

income of $990 from investments. What amount should be invested in the 7.5% bond?
Mathematics
1 answer:
nlexa [21]3 years ago
6 0

Answer:

$13,200

Step-by-step explanation:

You need to use the simple interest formula

I = P * r * t

I = Interest accrued

P = Principal amount invested

r = Interest rate          you need to divide by 100 to get it in decimal form

t = time, in years        if you are given a partial year, divide the months by 12

P = $12,000                                    

r = 7.5% = .075                                    

t = 1                                                  

But, because we want I to equal $990 then I is

I = $990

So we ignore our P and instead solve for the P that will give us the desired result.

I = P * r * t

$990 = P * .075 * 1

$990 = P.075        Divide each side by .075

$990/.075 = P.075/.075

$990/.075 = P

$13,200 = P

So, to earn an annual interest income of $990, $13,200 will have to be invested in the 7.5% bond.

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Part (c)

We'll use this identity

\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)\\\\

to say

\sin(A+45) = \sin(A)\cos(45) + \cos(A)\sin(45)\\\\\sin(A+45) = \sin(A)\frac{\sqrt{2}}{2} + \cos(A)\frac{\sqrt{2}}{2}\\\\\sin(A+45) = \frac{\sqrt{2}}{2}(\sin(A)+\cos(A))\\\\

Similarly,

\sin(A-45) = \sin(A + (-45))\\\\\sin(A-45) = \sin(A)\cos(-45) + \cos(A)\sin(-45)\\\\\sin(A-45) = \sin(A)\cos(45) - \cos(A)\sin(45)\\\\\sin(A-45) = \sin(A)\frac{\sqrt{2}}{2} - \cos(A)\frac{\sqrt{2}}{2}\\\\\sin(A-45) = \frac{\sqrt{2}}{2}(\sin(A)-\cos(A))\\\\

-------------------------

The key takeaways here are that

\sin(A+45) = \frac{\sqrt{2}}{2}(\sin(A)+\cos(A))\\\\\sin(A-45) = \frac{\sqrt{2}}{2}(\sin(A)-\cos(A))\\\\

Therefore,

2\sin(A+45)*\sin(A-45) = 2*\frac{\sqrt{2}}{2}(\sin(A)+\cos(A))*\frac{\sqrt{2}}{2}(\sin(A)-\cos(A))\\\\2\sin(A+45)*\sin(A-45) = 2*\left(\frac{\sqrt{2}}{2}\right)^2\left(\sin^2(A)-\cos^2(A)\right)\\\\2\sin(A+45)*\sin(A-45) = 2*\frac{2}{4}\left(\sin^2(A)-\cos^2(A)\right)\\\\2\sin(A+45)*\sin(A-45) = \sin^2(A)-\cos^2(A)\\\\

The identity is confirmed.

==========================================================

Part (d)

\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)\\\\\sin(45+A) = \sin(45)\cos(A) + \cos(45)\sin(A)\\\\\sin(45+A) = \frac{\sqrt{2}}{2}\cos(A) + \frac{\sqrt{2}}{2}\sin(A)\\\\\sin(45+A) = \frac{\sqrt{2}}{2}(\cos(A)+\sin(A))\\\\

Similarly,

\sin(45-A) = \sin(45 + (-A))\\\\\sin(45-A) = \sin(45)\cos(-A) + \cos(45)\sin(-A)\\\\\sin(45-A) = \sin(45)\cos(A) - \cos(45)\sin(A)\\\\\sin(45-A) = \frac{\sqrt{2}}{2}\cos(A) - \frac{\sqrt{2}}{2}\sin(A)\\\\\sin(45-A) = \frac{\sqrt{2}}{2}(\cos(A)-\sin(A))\\\\

-----------------

We'll square each equation

\sin(45+A) = \frac{\sqrt{2}}{2}(\cos(A)+\sin(A))\\\\\sin^2(45+A) = \left(\frac{\sqrt{2}}{2}(\cos(A)+\sin(A))\right)^2\\\\\sin^2(45+A) = \frac{1}{2}\left(\cos^2(A)+2\sin(A)\cos(A)+\sin^2(A)\right)\\\\\sin^2(45+A) = \frac{1}{2}\cos^2(A)+\frac{1}{2}*2\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\\sin^2(45+A) = \frac{1}{2}\cos^2(A)+\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\

and

\sin(45-A) = \frac{\sqrt{2}}{2}(\cos(A)-\sin(A))\\\\\sin^2(45-A) = \left(\frac{\sqrt{2}}{2}(\cos(A)-\sin(A))\right)^2\\\\\sin^2(45-A) = \frac{1}{2}\left(\cos^2(A)-2\sin(A)\cos(A)+\sin^2(A)\right)\\\\\sin^2(45-A) = \frac{1}{2}\cos^2(A)-\frac{1}{2}*2\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\\sin^2(45-A) = \frac{1}{2}\cos^2(A)-\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\

--------------------

Let's compare the results we got.

\sin^2(45+A) = \frac{1}{2}\cos^2(A)+\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\\sin^2(45-A) = \frac{1}{2}\cos^2(A)-\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\

Now if we add the terms straight down, we end up with \sin^2(45+A)+\sin^2(45-A) on the left side

As for the right side, the sin(A)cos(A) terms cancel out since they add to 0.

Also note how \frac{1}{2}\cos^2(A)+\frac{1}{2}\cos^2(A) = \cos^2(A) and similarly for the sin^2 terms as well.

The right hand side becomes \cos^2(A)+\sin^2(A) but that's always equal to 1 (pythagorean trig identity)

This confirms that \sin^2(45+A)+\sin^2(45-A) = 1 is an identity

4 0
3 years ago
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