005 Sample Final A, Question 14 - Revision history

Matt Lee at 04:10, 22 May 2015

2015-05-22T04:10:06Z

Kayla Murray at 20:14, 20 May 2015

2015-05-20T20:14:29Z

Kayla Murray at 19:20, 20 May 2015

2015-05-20T19:20:08Z

Matt Lee: Created page with "''' Question ''' Prove the following identity,
$\frac{1-\sin(\theta)}{\cos(\theta)}=\frac{\cos(\theta)}{1+\sin(\theta)}$ {| class="mw-colla..."

2015-05-01T04:18:44Z

Created page with "''' Question ''' Prove the following identity, <br> <center><math>\frac{1-\sin(\theta)}{\cos(\theta)}=\frac{\cos(\theta)}{1+\sin(\theta)}</math></center> {| class="mw-colla..."

New page

''' Question ''' Prove the following identity, <br>
<center><math>\frac{1-\sin(\theta)}{\cos(\theta)}=\frac{\cos(\theta)}{1+\sin(\theta)}</math></center>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Final Answers
|-
|a) False. Nothing in the definition of a geometric sequence requires the common ratio to be always positive. For example, <math>a_n = (-a)^n</math>
|-
|b) False. Linear systems only have a solution if the lines intersect. So y = x and y = x + 1 will never intersect because they are parallel.
|-
|c) False. <math>y = x^2</math> does not have an inverse.
|-
|d) True. <math>cos^2(x) - cos(x) = 0</math> has multiple solutions.
|-
|e) True.
|-
|f) False.
|}

← Older revision		Revision as of 04:10, 22 May 2015
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		+	{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"
		+	! Foundations:
		+	\|-
		+	\|1) What can you multiply <math>1- \sin(\theta)</math> by to obtain a formula that is equivalent to something involving <math>\cos</math>?
		+	\|-
		+	\|Answers:
		+	\|-
		+	\|1) You can multiply <math>1 - \sin(\theta)</math> by <math>\frac{1 + \sin(\theta)}{\1 + \sin(\theta)}</math>
		+	\|}

	{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"		{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 7: / Line 7: @@
 ! Step 1:
 |-
+|We start with the left hand side. We have <math>\frac{1-\sin(\theta)}{\cos(\theta)}=\frac{1-\sin(\theta)}{\cos(\theta)}\Bigg(\frac{1+\sin(\theta)}{1+\sin(\theta)}\Bigg)</math>.
 |}
@@ Line 13: / Line 13: @@
 ! Step 2:
 |-
+| Simplifying, we get <math>\frac{1-\sin(\theta)}{\cos(\theta)}=\frac{1-\sin^2(\theta)}{\cos(\theta)(1+\sin(\theta))}</math>.
 |}
@@ Line 19: / Line 19: @@
 ! Step 3:
 |-
+| Since <math>1-\sin^2(\theta)=\cos^2(\theta)</math>, we have
 |-
+|<math>\frac{1-\sin(\theta)}{\cos(\theta)}=\frac{\cos^2(\theta)}{\cos(\theta)(1+\sin(\theta))}=\frac{\cos(\theta)}{1+\sin(\theta)}</math>
 |}

@@ Line 5: / Line 5: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Final Answers
+! Step 1:
 |-
-|a) False. Nothing in the definition of a geometric sequence requires the common ratio to be always positive. For example, <math>a_n = (-a)^n</math>
 |-
-|b) False. Linear systems only have a solution if the lines intersect. So y = x and y = x + 1 will never intersect because they are parallel.
 |-
-|c) False. <math>y = x^2</math> does not have an inverse.
 |-
-|d) True. <math>cos^2(x) - cos(x) = 0</math> has multiple solutions.
 |-
-|e) True.
 |-
-|f) False.
 |}

005 Sample Final A, Question 14 - Revision history

Matt Lee at 04:10, 22 May 2015

Kayla Murray at 20:14, 20 May 2015

Kayla Murray at 19:20, 20 May 2015

Matt Lee: Created page with "''' Question ''' Prove the following identity, \frac{1-\sin(\theta)}{\cos(\theta)}=\frac{\cos(\theta)}{1+\sin(\theta)} {| class="mw-colla..."

Matt Lee: Created page with "''' Question ''' Prove the following identity,
$\frac{1-\sin(\theta)}{\cos(\theta)}=\frac{\cos(\theta)}{1+\sin(\theta)}$ {| class="mw-colla..."