Difference between revisions of "005 Sample Final A, Question 15"
Jump to navigation
Jump to search
Kayla Murray (talk | contribs) |
|||
(One intermediate revision by one other user not shown) | |||
Line 2: | Line 2: | ||
{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
− | ! | + | ! Foundations |
+ | |- | ||
+ | |1) <math>\tan^{-1}(x)</math> can be thought of as <math>\tan^{-1}\left(\frac{x}{1}\right),</math> and this now refers to an angle in a triangle. What are the side lengths of this triangle? | ||
|- | |- | ||
− | | | + | |Answers: |
− | |||
− | |||
− | |||
− | |||
|- | |- | ||
− | | | + | |1) The side lengths are 1, x, and <math>\sqrt{1 + x^2}.</math> |
|} | |} | ||
{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
− | ! Step | + | ! Step 1: |
|- | |- | ||
− | | | + | |First, let <math>\theta=\tan^{-1}(x)</math>. Then, <math>\tan(\theta)=x</math>. |
− | |||
− | |||
|} | |} | ||
{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
− | ! Step | + | ! Step 2: |
|- | |- | ||
− | | | + | | Now, we draw the right triangle corresponding to <math>\theta</math>. Two of the side lengths are 1 and x and the hypotenuse has length <math>\sqrt{x^2+1}</math>. |
|} | |} | ||
{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
− | ! Step | + | ! Step 3: |
|- | |- | ||
− | | | + | | Since <math>\cos(\theta)=\frac{\mathrm{opposite}}{\mathrm{hypotenuse}}</math>, <math>\cos(\tan^{-1}(x))=\cos(\theta)=\frac{1}{\sqrt{x^2+1}}</math>. |
|- | |- | ||
− | + | | | |
− | |||
− | | | ||
|} | |} | ||
Line 40: | Line 34: | ||
! Final Answer: | ! Final Answer: | ||
|- | |- | ||
− | | | + | | <math>\frac{1}{\sqrt{x^2+1}}</math> |
|} | |} |
Latest revision as of 21:13, 21 May 2015
Question Find an equivalent algebraic expression for the following,
Foundations |
---|
1) can be thought of as and this now refers to an angle in a triangle. What are the side lengths of this triangle? |
Answers: |
1) The side lengths are 1, x, and |
Step 1: |
---|
First, let . Then, . |
Step 2: |
---|
Now, we draw the right triangle corresponding to . Two of the side lengths are 1 and x and the hypotenuse has length . |
Step 3: |
---|
Since , . |
Final Answer: |
---|