Difference between revisions of "005 Sample Final A, Question 15"

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(Created page with "''' Question ''' Find an equivalent algebraic expression for the following, <center><math> \cos(\tan^{-1}(x))</math></center> {| class="mw-collapsible mw-collapsed" style = "...")
 
 
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! Final Answers
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! Foundations
 
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|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>
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|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?
 
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|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.
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|c) False. <math>y = x^2</math> does not have an inverse.
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|1) The side lengths are 1, x, and <math>\sqrt{1 + x^2}.</math>
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! Step 1:
 
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|d) True. <math>cos^2(x) - cos(x) = 0</math> has multiple solutions.
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|First, let <math>\theta=\tan^{-1}(x)</math>. Then, <math>\tan(\theta)=x</math>.
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! Step 2:
 
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|e) True.
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| 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>.
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! Step 3:
 
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|f) False.  
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| Since <math>\cos(\theta)=\frac{\mathrm{opposite}}{\mathrm{hypotenuse}}</math>, <math>\cos(\tan^{-1}(x))=\cos(\theta)=\frac{1}{\sqrt{x^2+1}}</math>.
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! Final Answer:
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| <math>\frac{1}{\sqrt{x^2+1}}</math>
 
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Latest revision as of 20:13, 21 May 2015

Question Find an equivalent algebraic expression for the following,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cos(\tan^{-1}(x))}
Foundations
1) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tan^{-1}(x)} can be thought of as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tan^{-1}\left(\frac{x}{1}\right),} 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{1 + x^2}.}
Step 1:
First, let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta=\tan^{-1}(x)} . Then, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tan(\theta)=x} .
Step 2:
Now, we draw the right triangle corresponding to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta} . Two of the side lengths are 1 and x and the hypotenuse has length Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{x^2+1}} .
Step 3:
Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cos(\theta)=\frac{\mathrm{opposite}}{\mathrm{hypotenuse}}} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cos(\tan^{-1}(x))=\cos(\theta)=\frac{1}{\sqrt{x^2+1}}} .
Final Answer:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{\sqrt{x^2+1}}}
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