<p>1) Find all solutions for </p>
<p>Tan^2 x = (sin x) / (1 + sin x)</p>
<p>2) Find x</p>
<p>½(3^x + 4 ∙ 3^-x) = 2 </p>
<p>3) Prove the identity.</p>
<p>(sin a cos a sin^2 b + cos^2 a tan a cos^2 b) csc a sec a = 1 </p>
<p>4) A cone has a height of 5 cm and a base with a diameter of 10 cm. A cylinder is placed in the center of the cone. Find the volume of the cylinder as a function of the radius of the cylinder. What is the domain of the function?
****For Questions 1 and 3 the exponents are 2. For question 2 the exponents are x and -x.</p>
<p>Since I won't be using sin(2x), cos(2x) nor tan(2x) here,
let sin2x = sin^2 x
let cos2x = cos^2 x
let tan2x = tan^2 x
and let sin2x / cos2x be understood as (sin^2 x)/(cos^2 x)
for the sake of ease of typing</p>
<p>1)
tan2x = sinx / (1+sinx)
sin2x / cos2x = sinx / (1 + sinx)
sin2x / (1 - sin2x) = sin / (1 + sinx) [because sin2x + cos2x = 1]
sin2x (1 + sinx) = sinx (1 - sin2x) [I didn't cancel so not to / by 0 ]
sin2x + sin3x = sinx - sin3x
2sin3x + sin2x - sinx = 0
sinx (2sinx + sinx - 1) = 0
sinx (sinx + 1)(2sinx - 1) = 0
let the domain be limited between 0 and 360 degrees
From sinx = 0, we get x = 0 or x = 180
From sinx + 1 = 0 => sinx = -1, we get x = 270
From 2sinx - 1 = 0, we get x = 30 or x = 150</p>
<p>2)
½(3^x + 4 ∙ 3^-x) = 2
3^x + 4 / 3^x = 4 [multiply both sides by 3^x]
3^2x + 4 = 4 ∙ 3^x
3^2x - 4 ∙ 3^x + 4 = 0 [let 3^x be y, to make things less confusing]
y^2 - 4y + 4 = 0
(y - 2)^2 = 0
y - 2 = 0 [sub 3^x back for y]
3^x - 2 = 0
3^x = 2<br>
x = log base 3 of 2</p>
<p>3)
(I'll use x for a and y for b, so that sina doesn't look like a word)
Required to Prove:
(sinx cosx sin2y + cos2x tanx cos2y) cscx secx = 1
(sinxcosx) (sin2y + cos2y) (1/sinx) (1/cosx) = 1 [common factor]
sinx cancels with 1/sinx, and cosx cancels with 1/cosx
leaves you with sin2y + cos2y = 1, which is equal by definition</p>
<p>4) I'm sorry, I don't understand this question. What does placing a cylinder at the center of a cone mean? Inscribe? If so how? Does the height equal to the radius?</p>
<p>Too bad I can't draw in here.
Could be prolly solved with similar triangles.
No time now.</p>
<p>Thanks Alot G_man 1988, you helped be a whole lot. The original problem for question number 4 goes like this. </p>
<p>A cone has a height of 5 cm and a base with a diameter 10 cm. A cylinder is placed in the center of the cone. Find the volume of the cylinder as a function of the radius of the cylinder. What is the domain of the function?
<strong><em>This is word for word of the original problem.</em></strong>***</p>
<p>Let h and r be the height and the radius of the cylinder(5-h)/5 = r/5 from the similar triangles
h=5-r
V=(pi)(r^2)(h)=(pi)(r^2)(5-r)
Romain 0<r<5
assuming the cylinder is inscribed in the cone which is the only thing that makes sense.</p>