Answer:
See below for answers and explanations
Step-by-step explanation:
Problem 1 (#3)
The formula for a regular octagon inscribed in a circle of radius [tex]r[/tex] is [tex]A=2\sqrt{2}r^2[/tex]. Hence, [tex]A=2\sqrt{2}(4)^2=16*2\sqrt{2}=32\sqrt{2}\approx45.255m^2[/tex].
Thus, C is the correct answer
Problem 2 (#4)
Using the co-function identity [tex]\displaystyle \sin\biggr(x+\frac{\pi}{2}\biggr)=\cos(x)[/tex], the equation can be rewritten as [tex]cos(x)=\frac{\sqrt{3}}{2}[/tex]. Using a unit circle, it's easy to see that the answer is [tex]\displaystyle\biggr\{\frac{\pi}{6},\frac{11\pi}{6}\biggr\}[/tex].
Thus, C is the correct answer
Problem 3
[tex]f(x)=g(x)\\\\2\sin^2 x-1=-\cos(x)\\\\2(1-\cos^2 x)-1=-\cos(x)\\\\2-2\cos^2 x-1=-\cos(x)\\\\1-2\cos^2 x=-\cos(x)\\\\2\cos^2x-\cos(x)-1=0[/tex]
Let [tex]u=\cos(x)[/tex], hence:
[tex]2u^2-u-1=0\\\\(2u+1)(u-1)=0[/tex]
[tex]\displaystyle2u+1=0\\\\2u=-1\\\\u=-\frac{1}{2}\\ \\\cos(x)=-\frac{1}{2}\\ \\x=\biggr\{\frac{2\pi}{3},\frac{4\pi}{3}\biggr\}[/tex]
[tex]u-1=0\\\\u=1\\\\\cos(x)=1\\\\x=\{0\}[/tex]
So, the solution set is [tex]\displaystyle\biggr\{0,\frac{2\pi}{3},\frac{4\pi}3}\biggr\}[/tex]
Thus, B is the correct answer
Problem 4 (#8)
If we construct a right triangle in Quadrant I with an opposite leg length of 1 unit and an adjacent leg length of 1 unit, this shows that [tex]\displaystyle\tan\theta=\frac{opposite}{adjacent}=\frac{1}{1}=1[/tex]. Since [tex]\displaystyle \csc\theta=\frac{1}{\sin\theta}[/tex] and [tex]\displaystyle\sin\theta=\frac{opposite}{hypotenuse}[/tex], we must solve for the hypotenuse with the Pythagorean Theorem:
[tex](opposite)^2+(adjacent)^2=(hypotenuse)^2\\1^2+1^2=(hypotenuse)^2\\1+1=(hypotenuse)^2\\2=(hypotenuse)^2\\\sqrt{2}=hypotenuse[/tex]
Therefore, since [tex]\displaystyle \sin\theta=\frac{opposite}{hypotenuse}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}[/tex], then [tex]\displaystyle \csc\theta=\frac{1}{\sin\theta}=\frac{1}{\frac{\sqrt{2}}{2}}=\frac{2}{\sqrt{2}}=\frac{2\sqrt{2}}{2}=\sqrt{2}[/tex].
Thus, B is the correct answer
Problem 5 (#9)
As no angles are given and only side lengths, we are forced to use the Law of Cosines to solve for the angles:
- Side "a" will be the distance from A to B with corresponding angle A
- Side "b" will be the distance from B to C with corresponding angle B
- Side "c" will be the distance from C to A with corresponding angle C
Angle A:
[tex]a^2=b^2+c^2-2bc\cos(A)\\400^2=500^2+600^2-2(500)(600)\cos(A)\\160000=250000+360000-600000\cos(A)\\160000=610000-600000\cos(A)\\-450000=-600000\cos(A)\\\frac{3}{4}=\cos(A)\\ A\approx41.410^\circ[/tex]
Angle B:
[tex]b^2=a^2+c^2-2ac\cos(B)\\500^2=400^2+600^2-2(400)(600)\cos(B)\\250000=160000+360000-480000\cos(B)\\250000=520000-480000\cos(B)\\-270000=-480000\cos(B)\\\frac{9}{16}=\cos(B)\\B\approx55.771^\circ[/tex]
Angle C:
By the Triangle Angle-Sum Theorem, [tex]C\approx82.819^\circ[/tex]
Hence, we can conclude that angle A is the smallest angle the swimmers must turn between the buoys.
Thus, C) 41.410° is the correct answer
Problem 6 (#unknown)
As we are given two angles and a side length, we can use the Law of Sines to find side length "b". Firstly, we need angle C so we can set up the proportion to find "b". By the Triangle Angle-Sum Theorem, [tex]m\angle C=60^\circ[/tex].
[tex]\frac{\sin(B)}{b}=\frac{\sin(C)}{c}\\\\\frac{\sin(64^\circ)}{b}=\frac{\sin(60^\circ)}{8}\\\\8\sin(64^\circ)=b\sin(60^\circ)\\\\b=\frac{8\sin(64^\circ)}{\sin(60^\circ)}\\ \\b\approx8.303[/tex]
Thus, B is the correct answer (how ironic lol)
Problem 7 (#13)
[tex]\displaystyle\sqrt{2}\cos2x=\sin^2x+\cos^2x\\\\\sqrt{2}\cos2x=1\\\\\cos2x=\frac{1}{\sqrt{2}}\\\\\cos2x=\frac{\sqrt{2}}{2}\\ \\2x=\frac{\pi}{4},\frac{7\pi}{4}\\ \\x=\frac{\pi}{8},\frac{7\pi}{8}[/tex]
Hence, B is the correct answer
Problem 8 (#14)
[tex]\displaystyle \sec\theta=2\\\\\frac{1}{\cos\theta}=2\\ \\\cos\theta=\frac{1}{2}\\ \\\theta=\frac{\pi}{3}+2\pi n,\frac{5\pi}{3}+2\pi n[/tex]
Thus, D is the correct answer
Problem 9 (#19)
The bottom graph looks correct as the period of the tangent function is [tex]\frac{\pi}{|b|}=\frac{\pi}{2}[/tex].
