ProvingTrigIdentities2

Proving Trig Identites using a Compound Problem

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This PG code shows how to have a multi-part question in which each part is revealed sequentially on its own html page.

• File location in OPL: FortLewis/Authoring/Templates/Trig/ProvingTrigIdentities2.pg

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PG problem file	Explanation
Problem tagging data	Problem tagging:
DOCUMENT(); loadMacros( "PGstandard.pl", "MathObjects.pl", "compoundProblem.pl", "Parser.pl", "PGunion.pl", ); TEXT(beginproblem()); BEGIN_PROBLEM();	Initialization: We use the compoundProblem.pl macro to generate a multi-part question in which each part is sequentially revealed on its own html page.
Context("Numeric")->variables->are(t=>"Real"); # # Redefine the sin(x) to be e^(pi x) # Context()->functions->remove("sin"); package NewFunc; # this next line makes the function a # function from reals to reals our @ISA = qw(Parser::Function::numeric); sub sin { shift; my $x = shift; return CORE::exp($x*3.1415926535); } package main; # Add the new functions to the Context Context()->functions->add(sin=>{class=>'NewFunc',TeX =>'\sin'}); $isProfessor = $studentLogin eq 'professor'; # # Set up the compound problem object. # $cp = new compoundProblem( parts => 3, totalAnswers => 3, parserValues => 1, allowReset => $isProfessor, ); $part = $cp->part;	Setup:
if ($part == 1) { BEGIN_TEXT ${BBOLD}Part 1 of 3:${EBOLD} $BR $BR ${BITALIC}Instructions:${EITALIC} You will need to submit your answers twice for each part. The first time you submit your answers they will be checked for correctness. When your answer is correct, check the box for ${BITALIC}Go on to next part${EITALIC} and click the submit button. You will not be able to go back to previous parts. $BR $BR In this multi-part problem, we will use algebra to verify the identity $BCENTER \( \displaystyle \frac{ \sin(t) }{ 1-\cos(t) } = \frac{ 1+\cos(t) }{ \sin(t) }. \) $ECENTER $BR First, using algebra we may rewrite the equation above as $BR $BR \( \displaystyle \sin(t) = \left( \frac{1+\cos(t)}{\sin(t)} \right) \cdot \Big( \) \{ ans_rule(20) \} \( \Big) \) END_TEXT ANS( Formula("1-cos(t)")->cmp() ); }	Part 1:
if ($part == 2) { BEGIN_TEXT ${BBOLD}Part 2 of 3:${EBOLD} $BR $BR Step 0: \( \displaystyle \frac{ \sin(t) }{ 1-\cos(t) } = \frac{ 1+\cos(t) }{ \sin(t) }. \) $BR $BR Step 1: \( \displaystyle \sin(t) = \left( \frac{1+\cos(t)}{\sin(t)} \right) \cdot ( 1 - \cos(t) ). \) $BR $HR $BR We may use algebra to rewrite the equation from Step 1 as $BR $BR \( \sin(t) \cdot \big( \) \{ ans_rule(20) \} \( \big) = \big(1+\cos(t)\big) \cdot \big(1-\cos(t)\big) \). END_TEXT ANS( Formula("sin(t)")->cmp() ); }	Part 2:
if ($part == 3) { BEGIN_TEXT ${BBOLD}Part 3 of 3:${EBOLD} $BR $BR Step 0: \( \displaystyle \frac{ \sin(t) }{ 1-\cos(t) } = \frac{ 1+\cos(t) }{ \sin(t) }. \) $BR $BR Step 1: \( \displaystyle \sin(t) = \left( \frac{1+\cos(t)}{\sin(t)} \right) \cdot ( 1 - \cos(t) ). \) $BR $BR Step 2: \( \displaystyle \sin(t) \sin(t) = (1+\cos(t))(1-\cos(t)) \) $BR $HR $BR Finally, using algebra we may rewrite the equation from step 2 as $BR $BR \( \sin^2(t) = \) \{ ans_rule(20) \} $BR $BR which is true since \( \cos^2(t) + \sin^2(t) = 1 \). Thus, the original identity can be derived by reversing these steps. END_TEXT ANS( Formula("1-(cos(t))^2")->cmp() ); } END_PROBLEM(); ENDDOCUMENT();	Part 3:

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