**QUESTION**

Trigonometry

Prove that:

a) (sin x + cos x)² + (sin x – cos x)² = 2

b) sin x / 1+cos x + 1+cos x / sin x = 2 / sin x

c) cos^4 x - sin^4 x + 1 = 2cos^2 x

d) (1-cos A) (1+1/cos A) = sin A tan A

**MY ANSWER**

a) sin^2 x + 2sin x cos x + cos^2 x

+

sin^2 x - 2sin x cos x + cos^2 x

=

2

Simplifying

sin^2 x + sin^2 x + cos^2 x + cos^2 x = 2

2sin^2 x + 2cos^2 x = 2

2(sin^2 x + 2cos^2 x) =2

2(1) = 2

2 = 2

b) multiply the equation by LCD = (1+cos x)(sin x)

this will result to

sin^2 x + (1 + cos x)^2 = 2(1 + cos x)

sin^2 x + 1 +2cos x + cos^2 x = 2 + 2cos x

sin^2 x + cos^2 x + 1 +2cos x = 2 + 2cos x

1 + 1 +2cos x = 2 + 2cos x

2 +2cos x = 2 + 2cos x

c) cos^4 x - sin^4 x + 1 = 2cos^2 x

Transpose cos^4 x to the other side

- sin^4 x + 1 = 2cos^2 x - cos^4 x

factor out cos^2 x

-sin^ 4 x + 1 = cos^2 x(2 - cos^2 x)

Breakdown 2 to 1 + 1

-sin^ 4 x + 1 = cos^2 x(1 + 1 - cos^2 x)

Use identities

-sin^4 x + 1 = cos^2 x (1 + sin^2 x)

Distribute

-sin^4 x+ 1 = cos^2 x + sin^2 x cos^2 x

Transpose

1 = cos^2 x + sin^2 x cos^2 x +sin^4 x

Factor out

1 = cos^2 x + sin^2 x(cos^2 x + sin^2 x)

Use identity

1 = cos^2 x + sin^2 x (1)

1 = (1)

d) Expand

1 + 1/cos A - cos A -1 = sin A tan A

Simplify and use identities

1/cos A - cos A = sin A sin A / cos A

multiply by LCD

1 - cos^2 A = sin^2 A

sin^2 A = sin^2 A

**ASKER'S COMMENT**

thanks so much with your working

**QUESTION**

if i were to borrow $25 from my dad, and $25 from mom, i would have $50 in my hand. i went out shopping to buy a t-shirt which cost me $45, means i have $5 left in my hand. i went home, decided to pay my debt little by little using the money i have left. since i have school the next day, i want to keep $3 to myself and pay dad $1 and mom $1. which means i owe them $24 each. $24x2=$48. but if this $48 is added with the $3 that i have in hand, it sums to $51. why is that so if i just owe them $50 in the first place??

**MY ANSWER**

Your analysis is wrong.

$24 * 2 = $48 <-- this should be negative because this is the amount of your debt. Add your $3 -$48 + $3 = -$45 What is the price of the pants you bought? $45-$45 = 0

**ASKER'S COMMENT**

thx for the clear answer!

**QUESTION**

tan1/2x + cot1/2x = 2cscx

how do i prove this?

**MY ANSWER**

Better write this in paper to be clearer

first Use identities to express the equation in terms of sine and cosine.

tan x = sin x/cos x

cot x = cos x/sin x

csc x = 1 / sin x

sin (x/2) / cos (x/2) + cos (x/2) / sin ( x/2) = 2 / sin x

combine into one fraction

[sin^2 (x/2) + cos^2 (x/2)] / sin (x/2) cos (x/2)= 2 /sin x

Manipulate and use again identities.

sin^2 x + cos^ 2 x = 1

2sin x cos x = sin 2x

[sin^2 (x/2) + cos^2 (x/2)] / 2sin (x/2) cos (x/2) = 1/sin x

1/sin x = 1/sin x

they are indeed the same

**ASKER'S COMMENT**

ty

**QUESTION**

Let P(t) represent the number of wolves in a population at time t years, when t≥0. The population P(t) is increasing at a rate directly proportional to 800 - P(t), where the constant of proportionality is k.

(a) If P(0) = 500 , find P(t) in terms of t and k

(b) If P(2) = 700 , find k.

(c) Find the lim of P(t) as t approaches infinity

....and if you can show how you got it that would be great.

**MY ANSWER**

Derive an equation from the problem.

dP/dt = k(800-P)

Apply variable separable.

dP/(800-P) = kdt

Intergrate both sides with limits.

it is stated that P(0) = 500 it means that if t=0, P =500.

We can use that as the Lower limits(LL).

We have unknown upper limits (UL). use P and t

⌡dP/(800-P)with UL(P) LL(500) = k⌡dt with UL(t) LL(0)

Remember upper limit - lower limit

-ln (800-P) + ln (800- 500) = k (t-0)

Manipulate so we can express the formula in terms of k and t:

Combine the natural log

ln [300/(800-P)] = kt

Raise both sides to eliminate the natural log

300/(800-P) = e^kt

Isolate P on one side.

P = 800 - (300/e^kt) <--- answer to a. --- b) just simply use in P the limits as 500 and 700 UL and LL respectively and in t use 0 and 2 UL and LL respectively. ⌡dP/(800-P)with UL(700) LL(500) = k⌡dt with UL(2) LL(0) like what i did earlier: -ln(800-700) + ln (800-500) = k (2-0) solving for k: k = {ln[(800-500)/(800-700)]} / 2 k = 0.5493 <--- answer to b ---- in c. just simply use infinity as the UL for t, and UL for P is P -ln(800-P) + ln(800-500) = 0.5493(inf) Evaluate: 300/(800-P) = e^inf 300/e^inf = 800 - P P=800 - 300/inf any number divided by infinity is zero: proof 1/0 = inf so... 1/inf =0 P = 800 <-- answer to c

**ASKER'S COMMENT**

Thanks so much! You laid out your steps very orderly and so it was easy to follow, and a big help!

**QUESTION**

jenny is solving the equation x^2 - 8x=9

by completing the square. what number should be added to both sides of the equation to complete the saquare?

what are the solutions to the equation? x^2 + 2x + 2=0

**MY ANSWER**

In completing the square we try to make the equation more like a perfect square trinomial.

A perfect square trinomial is the product of a square of a binomial.

To refresh you here are the steps to get the special product, square of a binomial.

(x+y)^2

1) Square the first term. x^2

2) Twice the product of the two terms. 2xy

3) Square the 2nd term. y^2

in x^2 - 8x= 9

our first goal is to make it a perfect square trinomial.

x^2 - 8x is already part of the trinomial, so we need one more, the last term.

Since the second term is twice the product of the two terms, we can compute for the last term.

1) Divide the linear x by two.

----> -8/2 = -4

2) Divide the result by the square root of the coefficient of the 1st term.

-----> -4/1 = -4

3) Square the result.

-----> -4^2 = 16

16 is the last term to make x^2 -8x a perfect square trinomial.

Next.

Add a zero to the equation.

You can do this by simply adding the last term and subtracting it off again.

---> x^2 -8x +16 -16 =9

Factor the perfect square trinomial and transpose all terms to one side then simplify:

----> (x-4)^2 -9 -16 = 0

----> (x-4)^2 -25= 0

The resulting equation is yet another special product. The difference of two squares.

This special product is the product of the sum and difference of two terms.

(x+y)(x-y)

Steps:

1) Square the 1st term. x^2

2) Place a negative sign. -

3) Square the second term. y^2

as you can see:

(x-4)^2 is a square

- then we have a negative sign

25 is a perfect square.

So factor it.

Steps:

1) Get the square root of both terms

---> sqrt(x^2) & sqrt(y^2) = x y

2) Place plus and minus sign inbetween them.

---> (x+y)(x-y)

Result:

[(x-4)+5] [ (x-4)-5] = 0.

Solve for x:

[(x-4)+5] = 0

x=-1

[ (x-4)-5]

x =9

-----

2nd equation is imaginary.

use quadratic formula.

[-b +- sqrt(b^2 - 4ac) ]/2a

in the form

ax^2 + bx + c =0

the discriminant (b^2 - 4ac) will be negative

2^2-4(1)(2)

4-8 = -4

square root of negative numbers are imaginary

**ASKER'S COMMENT**

thanks

**QUESTION**

1. A lawyer bills her clients $250 per hour of service. If a client's case requires 47 hours to complete, use proportion to calculate how much the client will owe the lawyer (excluding tax).

2. A new virus is released on the internet; the administrator of a department's Local Area Network ( LAN) is given five minutes by a manager to estimate the impact. The administrator samples 12 of the PCs connected to the LAN, and finds that 7 are infected; use proportion to estimate the number of infected PCs if there are a total of 117 PCs connected to the LAN.

**MY ANSWER**

its easier when using the units as guide.

1) $250/hr to be completed in 47 hrs.

Required answer in dollars

To cancel the unit hr, simply multip[ly the two

$250/hr * 47 hrs. = $11750

2) ration and proportion

7 is the representation of the infected PCs

12 is the representation of all the PCs

117 is the total number of PCs

x is the total number of infected PCs

7/12 = x/117

solve for x

x = 7*117/12

x= 68.25 ~ 68PCs are infected

**ASKER'S COMMENT**

I still got points taken off for #1; my teacher said that still isn't an equation. But thank you for trying to help.

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