Mathematics, 05.02.2020 12:52 keenonwhite2612

Suppose (theta sign) is an angle in standard position whose terminal side lies in the given quadrant. for these functions, find the exact values of the remaining five trigonometric funtctions of (theta)

sin(theta)=-4/5; quadrant 4

tan(theta)=2; quadrant 1

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_______________

A) $\mathsf{sin\,\theta=-\,\dfrac{4}{5}\qquad\qquad\theta \in 4th~quadrant.}$

$\mathsf{5\,sin\,\theta=-4\qquad\quad(i)}$

• Finding $\mathsf{cos\,\theta:}$

Square both sides of $\mathsf{(i):}$

$\mathsf{(5\,sin^2\,\theta)=4^2}\\\\
\mathsf{5^2\,sin^2\,\theta=4^2}\\\\
\mathsf{25\,sin^2\,\theta=16\qquad\qquad(but~sin^2\,\theta=1-cos^2\,\theta)}\\\\
\mathsf{25\cdot (1-cos^2\,\theta)=16}\\\\
\mathsf{25-25\,cos^2\,\theta=16}$

$\mathsf{25-16=25\,cos^2\,\theta}\\\\
\mathsf{9=25\,cos^2\,\theta}\\\\
\mathsf{cos^2\,\theta=\dfrac{9}{25}}\\\\\\
\mathsf{cos\,\theta=\pm\,\sqrt{\dfrac{9}{25}}}\\\\\\
\mathsf{cos\,\theta=\pm\,\sqrt{\dfrac{3^2}{5^2}}}\\\\\\
\mathsf{cos\,\theta=\pm\,\dfrac{3}{5}}$

But $\mathsf{cos\,\theta}$ is also negative, because $\mathsf{\theta}$ lies in the 4th quadrant. So,

$\mathsf{cos\,\theta=-\,\dfrac{3}{5}\qquad\quad\checkmark}$

• Finding $\mathsf{tan\,\theta:}$

$\mathsf{tan\,\theta=\dfrac{sin\,\theta}{cos\,\theta}}\\\\\\
\mathsf{tan\,\theta=\dfrac{-\,\frac{4}{5}}{-\,\frac{3}{5}}}\\\\\\
\mathsf{tan\,\theta=-\,\dfrac{4}{\diagup\hspace{-6}5}\cdot \left(\!-\,\dfrac{\diagup\hspace{-6}5}{3}\right)}\\\\\\
\mathsf{tan\,\theta=\dfrac{4}{3}\qquad\quad\checkmark}$

• Finding $\mathsf{cot\,\theta:}$

$\mathsf{cot\,\theta=\dfrac{1}{tan\,\theta}}\\\\\\
\mathsf{cot\,\theta=\dfrac{1}{~\frac{4}{3}~}}\\\\\\
\mathsf{cot\,\theta=\dfrac{3}{4}\qquad\quad\checkmark}$

• Finding $\mathsf{sec\,\theta:}$

$\mathsf{sec\,\theta=\dfrac{1}{cos\,\theta}}\\\\\\
\mathsf{sec\,\theta=\dfrac{1}{-\,\frac{3}{5}}}\\\\\\
\mathsf{sec\,\theta=-\,\dfrac{5}{3}\qquad\quad\checkmark}$

• Finding $\mathsf{csc\,\theta:}$

$\mathsf{csc\,\theta=\dfrac{1}{sin\,\theta}}\\\\\\
\mathsf{csc\,\theta=\dfrac{1}{-\,\frac{4}{5}}}\\\\\\
\mathsf{csc\,\theta=-\,\dfrac{5}{4}\qquad\quad\checkmark}$

________

B) $\mathsf{tan\,\theta=2\qquad\qquad\theta \in 1st~quadrant.}$

$\mathsf{\dfrac{sin\,\theta}{cos\,\theta}=2}\\\\\\
\mathsf{sin\,\theta=2\,cos\,\theta\qquad\quad(ii)}$

• Finding $\mathsf{sin\,\theta:}$

Square both sides of $\mathsf{(ii):}$

$\mathsf{(sin\,\theta)^2=(2\,cos\,\theta)^2}\\\\
\mathsf{sin^2\,\theta=2^2\,cos^2\,\theta}\\\\
\mathsf{sin^2\,\theta=4\,cos^2\,\theta\qquad\qquad(but~cos^2\,\theta=1-sin^2\,\theta)}\\\\
\mathsf{sin^2\,\theta=4\cdot (1-sin^2\,\theta)}\\\\
\mathsf{sin^2\,\theta=4-4\,sin^2\,\theta}$

$\mathsf{sin^2\,\theta+4\,sin^2\,\theta=4}\\\\
\mathsf{5\,sin^2\,\theta=4}\\\\
\mathsf{sin^2\,\theta=\dfrac{4}{5}}\\\\
\mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{4}{5}}}\\\\\\
\mathsf{sin\,\theta=\pm\,\dfrac{2}{\sqrt{5}}}$

But sine is positive because $\theta$ is a 1st quadrant angle:

$\mathsf{sin\,\theta=\dfrac{2}{\sqrt{5}}\qquad\quad\checkmark}$

• Finding $\mathsf{cos\,\theta:}$

$\mathsf{sin\,\theta=2\,cos\,\theta}\\\\
\mathsf{cos\,\theta=\dfrac{1}{2}\,sin\,\theta}\\\\\\
\mathsf{cos\,\theta=\dfrac{1}{\diagup\hspace{-6}2}\cdot \dfrac{\diagup\hspace{-6}2}{\sqrt{5}}}\\\\\\
\mathsf{cos\,\theta=\dfrac{1}{\sqrt{5}}\qquad\quad\checkmark}$

• Finding $\mathsf{cot\,\theta:}$

$\mathsf{cot\,\theta=\dfrac{1}{tan\,\theta}}\\\\\\
\mathsf{cot\,\theta=\dfrac{1}{2}}\qquad\quad\checkmark}$

• Finding $\mathsf{sec\,\theta:}$

$\mathsf{sec\,\theta=\dfrac{1}{cos\,\theta}}\\\\\\ \mathsf{sec\,\theta=\dfrac{1}{~\frac{1}{\sqrt{5}}~}}\\\\\\ \mathsf{sec\,\theta=\sqrt{5}\qquad\quad\checkmark}$

• Finding $\mathsf{csc\,\theta:}$

$\mathsf{csc\,\theta=\dfrac{1}{sin\,\theta}}\\\\\\
\mathsf{csc\,\theta=\dfrac{1}{~\frac{2}{\sqrt{5}}~}}\\\\\\
\mathsf{csc\,\theta=\dfrac{\sqrt{5}}{2}\qquad\quad\checkmark}$

I hope this helps. =)

Tags: trigonometric trig function sine cosine tangent cotangent secant cosecant sin cos tan cot sec csc trigonometry

Answer from: Quest

part a

answer: a reasonable domain is $0 \le n \le 10$ where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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

part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

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

part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).

Answer from: Quest

none of these

step-by-step explanation:

23 + 34 add up to 57 so just inform the teacher of this really stupid mistake

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