Answer to Question #159312 in Calculus for Jon jay Mendoza

Question

1.   Verify the identity 1 - sin² x cot² x = sin² x. Is there more than one way to verify the identity? If so, tell which way you think is easier and why. 

2.John said 𝒔𝒊𝒏 𝒙 + 𝒄𝒐𝒔 𝒙 = 𝟐 has no solution. Do you agree with John? Explain why or why not? 

3.

SKILL CHECK: Verify each identity.  

1.𝒄𝒐𝒕² 𝜽+1 / cot²𝜽=sec²𝜽

2.(𝒄𝒔𝒄^𝟐𝜽 − 𝟏)𝒔𝒊𝒏^𝟐𝜽 ≡ 𝒄𝒐𝒔^𝟐𝜽 

3.8. 𝟏 − 𝒔𝒆𝒄 𝜶 𝒄𝒐𝒔 𝜶 ≡ 𝒕𝒂𝒏 𝜶 𝒄𝒐𝒕 𝜶 − 𝟏 

4.tan A + cot A / sec A csc A =1

5.𝟏 + 𝟐 𝒕𝒂𝒏 ^𝟐𝜽 ≡ 𝒔𝒆𝒄^𝟒𝜽 − 𝒕𝒂𝒏 ^𝟒𝜽 

Trigonometric equation

6.cos x + √𝟑 = - cos x 

7.sin2 x – tan x cos x = 0

8.sin x + √𝟐 = − sin x 

9. 2 cos2 x – 5 cos x = 3

10.√𝟑 csc x + 2 = 0

solve the worded problem

1. On which days of the year are there 10 hours of sunlight in Prescott, Arizona?

2.The tide, or depth of the ocean near the shore, changes throughout the day. The depth of the Bay of Fundy can be modeled by....... where d is the water depth in feet and t is the time in hours. Consider a day in which t = 0 represents 12:00 A.M. At what time(s) is the water depth 3 1 2 feet?

Answer 1

$\displaystyle 1).\\ 1 - \sin^2{x}\cot^2{x} \\ \textsf{There is more than one way} \\ \begin{aligned} 1 - \sin^2{x}\cot^2{x} &= 1 - \sin^2{x}\cdot\frac{cos^2{x}}{\sin^2{x}}\\ &= 1 - \cos^2{x} = \sin^2{x} \end{aligned} \\ \begin{aligned} \cos^2{x} + \sin^2{x} &= 1\\ \cot^2{x} + 1 &= \cosec^2{x} \\ \cot^2{x} &= \cosec^2{x} - 1 \end{aligned} \\ \begin{aligned} 1 - \sin^2{x}\cot^2{x} &= 1 - \sin^2{x}\left(\cosec^2{x} - 1\right)\\ &= 1 - 1 + \sin^2{x} = \sin^2{x} \end{aligned}\\ \textsf{The first way is easier because it}\\ \textsf{involves a short process and}\\ \textsf{is straightfoward.}\\ 2).\\ \sin^2{x} + \cos^2{x} = 2 \\ \textsf{It has no solution.} \textsf{The maximum value of}\\ \cos^2{x}\,\, \textsf{is}\,\, 1 \,\, \textsf{at}\,\, x = 0.\\ \textsf{The maximum value of}\,\, \sin^2{x}\\ \textsf{is}\,\, 1\,\, \textsf{at}\,\, x = \frac{\pi}{2}.\\ \textsf{The values of}\,\, x\,\, \textsf{do not coincide}\\ \textsf{at the maximum values and as such,}\\ \textsf{there is no solution to the given equation.}\\ 3).\\ \begin{aligned} \frac{\cot^2{\theta} + 1}{\cot^2{\theta}} &= 1 + \tan^2{\theta} \\&= \sec^2{\theta} \end{aligned} \\ 4).\\ \begin{aligned} (\cosec^2{\theta} - 1)\sin^2{\theta} &= 1 - \sin^2{\theta} \\&= \cos^2{\theta} \end{aligned} \\ 5).\\ \begin{aligned} 1 - \sec\alpha\cos\alpha &= 1- \frac{\cos\alpha}{\cos\alpha} \\&= 1 - 1 = 0 \end{aligned} \\ \begin{aligned} \tan\alpha\cot\alpha - 1 &= \frac{\tan\alpha}{\tan\alpha} - 1 \\&= 1 - 1 = 0 \end{aligned} \\ \therefore 1 - \sec\alpha\cos\alpha =\tan\alpha\cot\alpha - 1\\ 6).\\ \begin{aligned} \frac{\tan A + \cot A}{\sec A \cosec A} &= (\tan A + \cot A)\sin A \cos A \\&= \left(\frac{\sin A}{\cos A}+ \frac{\cos A}{\sin A}\right)\sin A \cos A = \\&= \sin^2 A + \cos^2 A = 1 \end{aligned} \\ 7).\\ 1 + 2\tan^2{\theta} = \sec^2{\theta} + \tan^2{\theta}\\ \textsf{It is known that}\,\, \cos^2{\theta} + \sin^2{\theta} = 1\\ \tan^2{\theta} + 1 = \sec^2{\theta}\\ \sec^2{\theta} - \tan^2{\theta}\\ \begin{aligned} 1 + 2\tan^2{\theta} &= \sec^2{\theta} + \tan^2{\theta} \\&= (\sec^2{\theta} + \tan^2{\theta})(1) = (\sec^2{\theta} + \tan^2{\theta})(\sec^2{\theta} - \tan^2{\theta}) \\&= \sec^4{\theta} - \tan^4{\theta} \end{aligned} \\ 8).\\ \cos{x} + \sqrt{3} = -\cos{x}\\ 2\cos{x} = \sqrt{3}\\ \cos{x} = \frac{\sqrt{3}}{2},\,\, x = \frac{\pi}{6}\\ 9).\\ \sin(2x) - \tan{x}\cos{x} = 0\\ 2\sin{x}\cos{x} - \sin{x} = 0\\ \sin{x}(2\cos{x} - 1) = 0\\ \sin{x} = 0, x = n\pi\\ 2\cos{x} - 1 = 0\\ \cos{x} = \frac{1}{2}\\ x = \frac{\pi}{6} + 2n\pi\\ \therefore x = n\pi,\,\, \frac{\pi}{6} + 2n\pi \,\, \forall n \in \mathbb{Z}\\ 10).\\ \sin{x} + \sqrt{2} = - \sin{x}\\ 2\sin{x} = \sqrt{2}\\ \sin{x} = \frac{\sqrt{2}}{2},\,\, x = \frac{\pi}{4}\\ 11).\\ 2\cos^2{x} - 5\cos{x} = 3\\ 2\cos^2{x} - 5\cos{x} - 3 = 0\\ \cos{x} = \frac{5 \pm 7}{4} \\ \cos{x} = 3,\,\, \cos{x} = -\frac{1}{2}\\ \cos(\pi + x) = \cos\left(\frac{\pi}{3}\right)\\ x = -\frac{2\pi}{3} + 2n\pi \,\, \forall n \in \mathbb{Z}\\ \cos{x} = \cos\left(\frac{2\pi}{3}\right) \\ x = \frac{2\pi}{3} + 2n\pi \,\, \forall n \in \mathbb{Z}\\ 12).\\ \sqrt{3}\cosec{x} + 2 = 0 \\ \cosec{x} = \frac{-2}{\sqrt{3}} \\ \sin{x} = -\frac{\sqrt{3}}{2} \\ \sin(\pi + x) = \frac{\sqrt{3}}{2} = \sin\left(\frac{\pi}{3}\right) \\ x = -\frac{2\pi}{3} + 2n\pi \,\, \forall n \in \mathbb{Z} \\ \sin(2\pi - x) = \frac{\sqrt{3}}{2} = \sin\left(\frac{\pi}{3}\right) \\ x = \frac{5\pi}{3} + 2n\pi \,\, \forall n \in \mathbb{Z}$