7.7+OFSA+Solutions


 * OFSA SOLUTIONS for Verifying & Simplifying Using Trigonometric IDs**

This page contains peer generated solutions and error explanations to OFSA questions. As you read or view the solutions, be critical: check for accuracy, but also for more efficient solution strategies. If you have a better method or different idea/answer, post a discussion and monitor the responses.
 * Quick Directions**
 * Post answers, solutions and error explanations to each OFSA question below.
 * For each "distractor" or incorrect answer choice, explain the error that would lead to that incorrect answer choice.
 * You may either do the above in typed format or using a pencast.
 * Separate each question with a section bar.
 * After each solution, provide a hyperlink back to the corresponding OFSA page.
 * Follow example below.

math {\text{Which one expression below is equivalent to: }}\sin \left( {x - \frac{\pi }{6}} \right) + \sin \left( {x + \frac{\pi }{6}} \right)? math
 * Sample Question**

math \begin{gathered} {\text{ }}\sin \left( {x - \frac{\pi }{6}} \right) - \sin \left( {x + \frac{\pi }{6}} \right) = \hfill \\ {\text{ }}\left( {\sin x\cos \frac{\pi }{6} - \cos x\sin \frac{\pi }{6}} \right) - \left( {\sin x\cos \frac{\pi }{6} + \cos x\sin \frac{\pi }{6}} \right) = \hfill \\ {\text{ }}\sin x \cdot \frac{2} - \cos x \cdot \frac{1}{2} - \sin x \cdot \frac{2} - \cos x \cdot \frac{1}{2} = \hfill \\ {\text{ }}\boxed{ - \cos x} \hfill \\ \end{gathered} math
 * __Solution__**

math \begin{array}{*{20}{l}} &{\text{0}}& \\ &{ - \cos x}& \\ &{\cos x}& \\ &{ - \sqrt 3 \cos x}&{{\text{Error. Student used incorrect unit circle values for sin}}\frac{\pi }{6}{\text{and cos}}\frac{\pi }{6}.} \end{array} math
 * __Error Explanation__**

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math {\text{Find the exact value of: }} \cos(75^{\circ}) math
 * Question 1**

math \begin{gathered} {\text{ }} \cos(45^{\circ}+30^{\circ}) = \hfill \\ {\text{ }} \cos45^{\circ}\cos30^{\circ}-\sin45^{\circ}\sin30^{\circ} = \hfill \\ {\text{ }} \frac{\sqrt2}{2}(\frac{\sqrt3}{2}) - \frac{\sqrt2}{2}(\frac{1}{2}) = \hfill \\ {\text{ }} \boxed{\frac{\sqrt6 - \sqrt 2} {4}} \hfill \\ \end{gathered} math
 * __Solution 1__**

__**Error Explanation 1**__ math \begin{array}{*{20}{l}} &{\frac{-\sqrt 6-\sqrt2}{4}& \\ &{\frac{\sqrt 6-\sqrt2}{4} & \\ &{\frac{\sqrt 6+\sqrt2}{4} & {{ \text{Error. Student thought it was }} {\cos A+B = \cos a\cos b + \sin a\sin b}. \\ &{\frac{-\sqrt 6+\sqrt2}{2} & {{ \text{Error. Student made it }} {\sin a\sin b - \cos a\cos b } \\ \end{array} math

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math {\text{Which is equal to: }} 2\sin \frac{\Pi}{6} \cos \frac{\Pi}{6} math math \begin{gathered} {\text{ }} \sin (2\frac{\Pi}{6}) = \hfill \\ {\text{ }} \sin (\frac{\Pi}{3}) = \hfill \\ {\text{ }} \boxed{\frac{\sqrt3}{2}} \hfill \\ \end{gathered} math
 * Question 2**
 * __Solution 2__**

__**Error Explanation 2**__ math \begin{array}{*{20}{l}} &{\sqrt 3}& \\ &{\frac{1}{2}}& {\sin a}  {\sin 2a}. \\ &{\sin^{-1}\frac{\Pi}{12}}& { \sin \frac{1}{2}a} {\sin 2a}. \\ &{\frac{\sqrt3}{2}}&{{\text{Correct Answer.} \\ \end{array} math

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math 4\cos^{2}\frac{\pi}{6} - 2 math math \begin{array}{*{20}{l}} 2(2\cos^{2}\frac{\pi}{6} - 1) = \\ 2(\cos\frac{\pi}{3}) = \\ 2(\frac{1}{2}) = \\ 1 \end{array} math math \begin{array}{*{20}{l}} {\text A.} \; \sqrt{3} ;\ {\text {Error. Student used wrong cosine value}} \\ {\text B.} \; \; {\text Correct} \\ {\text C.} \; -\sqrt{3} ;\ {\text {Error. Student used wrong cosine value and not negative}} \\ {\text D.} \; -1 \; {\text {Error. Value is not negative}} \end{array} math
 * Question 3**
 * __Solution 3__**
 * __Error Explanation 3__**

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math {\text{Simplify: }} \left ( \tan^{2}x+1 \right )\left ( cos^{2}(-x)-1 \right )-tan^{2}(-x) math math \begin{gathered} \left (\sec^{2}x\right )\left ( \cos^{2}x-1 \right )+\tan^{2}x=\\ \left ( \frac{1}{\cos^{2}x} \right )\left (-\sin^{2}x \right )+\tan^{2}x=\\ -\left (\frac{\sin^{2}x}{\cos^{2}x} \right )+\tan^{2}x=\\ -\tan^{2}x+\tan^{2}x=\\ {\text{ }}\boxed{ 0}\hfill \\ \end{gathered} math math \begin{gathered} A.\ 2\tan^{2}x\ \textup{Error: Forgot to make sine term negative}\\ B.\ 0\ \textup{Correct!}\\ C.\ -2\tan^{2}x\ \textup{Error: Used incorrect even-odd identity}\\ D.\ -1+\tan^{2}x\ \textup{Error: Used incorrect reciprocal identity} \\ \end{gathered} math Click here to return to OFSA.
 * Question 4**
 * __Solution 4__**
 * __Error Explanation 4__**

math {\text {Simplify: }} \; \frac{(\sin x\cos x)^{2}}{\sin x\cos x} \\ \\ math math \frac{\sin^{2}x + 2\sin x\cos x + \cos^{2}x}{\sin x \cos x} = \\ \\ \frac{1 + 2\sin x \cos x}{\sin x \cos x} \\ \\ \frac{1}{\sin x \cos x} + \frac{2\sin x \cos x}{\sin x \cos x} = \\ \\ (\sec x \csc x) + 1\\ math math {\text {A.} \sec x\csc x - 2} \; {\text {Error. Used the wrong sign, not -2.}}\\
 * Question 5**
 * __Solution 5__**
 * __Error Explanation 5__**

{\text {B.} \sec x\csc x +2} \; {\text {Correct}}\\

{\text {C.} \sec x\csc x - 1} \; {\text {Error. Did not know how to simplify or did not know how to distribute at the beginning}} \\

{\text {D.} \sec x\csc x + 1} \; {\text {Error. Did not know how to simplify or did not know how to distribute at the beginning}} \\ math

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math {\text {Evaluate:}} \; \frac{11\pi}{3} math math {\text {Reference angle for}} \; \frac{11\pi}{3} \; is \; \frac{\pi}{3} \\ \tan \frac{11\pi}{3} = \\ -\frac{\sin \frac{\pi}{3}}{\cos \frac{\pi}{3}} = \\ -\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \\ -\sqrt{3} math math \begin{array}{*{20}{l}} {\text {A.}} \; \; -\sqrt{3} \;\; {\text {Correct Answer}}\\ {\text {B.}} \; \; \; \sqrt{3} \;\; Error. \; Student \; used \; positive \; tangent \; value. \\ {\text {C.}} \; \; -\frac{\sqrt{3}}{3} \;\; Error. \; Student \; came \; up \; with \; wrong \; tangent \; value. \\ {\text {D.}} \; \; \; \frac{\sqrt{3}}{3} \;\; Error. \; Student \; came \; up \; with \; wrong \; tangent \; value \; and \; negative \; value. \\ \end{array} math
 * Question 6**
 * __Solution 6__**
 * __Error Explanation 6__**

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math \[Evaluate: sin(45)cos(30)+cos(45)(sin30)\] math Use the idea of double angles. math \[sin(A+B)\] math math \[sin(45+30)=sin(75)\] math
 * Question 7**
 * __Solution 7__**
 * Solution is B.**

math \[sin(15)\neq sin(A+B)\] math math \[sin(\frac{\sqrt{2}}{2}*\frac{\sqrt{3}}{2})-(\frac{\sqrt{2}}{2}*\frac{1}{2})= sin\frac{\sqrt{6}-\sqrt{2}}{4}\] math math \[\frac{\sqrt{6}-\sqrt{2}}{4}\neq sin(A+B)\] math Wrong values on the unit circle. Reminder to study over the unit circle values. Click here to return to OFSA.
 * __Error Explanation 7__**
 * A is incorrect.**
 * C is incorrect.**
 * D is incorrect.**

math \[Evaluate: Tan(\frac{\sqrt{3}}{3})\] math Remember the domain of Tan math \[(\frac{-\Pi }{2}, \frac{\Pi }{2})\] math Remember which quadrants have positive tangent values (Quad I and III) negative tangent values (Quad II and IV) math \[D. \frac{\Pi }{6}\] math Only answer in the correct domain and quadrant for the given tangent value.
 * Question 8**
 * __Solution 8__**
 * Solution is D.**

math \[tan(\frac{\sqrt{3}}{3}) = \frac{\Pi }{6}, \frac{7\Pi }{6}\] math math \[tan(\frac{\sqrt{3}}{3}) \neq Tan(\frac{\sqrt{3}}{3})\] math Cannot be in quadrant III because of the domain. Incorrect quadrants. Remember which tangent values are positive and which are negative. Avoidable mistake (: Domain error with Tan. Click here to return to OFSA.
 * __Error Explanation 8__**
 * Solution A is incorrect.**
 * Solution B is incorrect.**
 * Solution C is incorrect.**

math \[Evaluate: cos(-x)+sin(x)\] math math \[Evaluate: cos(-x)+sin(x)\] math math \[cos(x)+sin(x)\] math __Even/Odd Identities__ math \[cos^{2}(x)+sin^{2}(x)\] math __Pythagorean Identities__ math \[cos^{2}(x)+sin^{2}(x)= 1\] math
 * Question 9**
 * __Solution 9__**
 * Solution B is correct.**

math \[cos(-x)\neq -cos(x)\] math Remember the EOIDs. No ID's support this idea. http://pch-wiki.wikispaces.com/7.1+Content+-+Verifying+%26+Simplifying+Using+Trigonometric+IDs With the correct ID's, it can be proven. http://pch-wiki.wikispaces.com/7.1+Content+-+Verifying+%26+Simplifying+Using+Trigonometric+IDs Click here to return to OFSA.
 * __Error Explanation 9__**
 * Solution A is incorrect.**
 * Solution C is incorrect.**
 * Solution D is incorrect.**

Solve: math (cos (5\pi/12))(cos (\pi/12)) + (sin (5\pi/12))(sin (\pi/12)) math
 * Question 10**

math cos(5\pi/12 - \pi/12)= math math cos(4\pi/12)= math math cos(\pi/3)= math math {\text{ }}\boxed{1/2} \hfill \\ \end{gathered} math
 * __Solution 10__**

math \begin{array}{*{20}{l}} {\text A.} \; \frac{1}{2};\ {\text {Correct!}} \\ {\text B.} \; {0} \; {\text {Error. Student thought the sum and difference formula was cos(A+B)=cosAcosB+sinAsinB}} \\ {\text C.} \; -\frac{\sqrt{3}}{2}};\ {\text {Error. Student thought the sum and difference formula for cosine was sin(A-B)=cosAcosB+sinAsinB}} \\ {\text D.} \; 1 \; {\text {Error. Student thought the sum and difference formula for cosine was sin(A+B)= cosAcosB+sinAsinB}} \end{array} math
 * __Error Explanation 10__**

math \text{A. } \frac{1}{2} \\ math

math \text{B. } 0 \\ math

math \text{C.} \frac{\sqrt{3}}{2}} \\ math

math \text{D. } 1 \\ math Click here to return to OFSA.

math \textup{Evaluate:}\ \left (1-\cos x\right )\left ( \csc x+\cot x \right )\\ math math \begin{gathered} \left ( 1-\cos x \right )\left ( \frac{1}{\sin x}+\frac{\cos x}{\sin x} \right )\\ \left (1-\cos x \right )\left (\frac{1+\cos x}{\sin x} \right )\\ \frac{1-\cos^{2}x}{\sin x }\\ \frac{\sin^{2}x}{\sin x}\\ {\text{ }}\boxed{\sin x}\hfill \\ \end{gathered} math math \textup{A.}\ \sin x\ \textup{Correct!}\\ \textup{B.}\ \frac{1-\cos x}{\sin x}\ \textup{Error: Incorrectly multiplied difference of squares.}\\ \textup{C.}\ \frac{\cot^{2} x}{\sin x}\ \textup{Error: Use of incorrect PID.}\\ \textup{D.}\ \textsl{Cannot be evaluated.}\ \textup{Error: Clearly, this CAN be evaluated!} math Click here to return to OFSA.
 * Question 11**
 * __Solution 11__**
 * __Error Explanation 11__**

math \textup{Simplify:}\ \frac{\sin x}{1+\cos x}+\frac{1+\cos x}{\sin x}\\ math math \begin{gathered} \frac{\sin^{2}x+\left ( 1+\cos x \right )^{2}}{\sin x\left ( 1+\cos x \right )}\\ \frac{\sin^{2}x+1+2\cos x+\cos^{2}x}{\sin x\left ( 1+\cos x \right )}\\ \frac{1+1+2\cos x}{\sin x\left ( 1+\cos x \right )}\\ \frac{2\left ( 1+\cos x\right )}{\sin x\left ( 1+\cos x \right )}\\ \frac{2}{\sin x}\\ {\text{ }}\boxed{2\csc x}\hfill \\ \end{gathered} math math \begin{gathered} \textup{A.}\ \frac{2\cos x}{\sin x\left ( 1+\cos x \right )}\ \textup{Error: Use of incorrect PID}\\ \textup{B.}\ 2\sec x\ \textup{Error: Use of incorrect RID}\\ \textup{C.}\ \sin x+1+\cos x\ \textup{Error: Canceled terms on numerator and denominator even though numerator is addition}\\ \textup{D.}\ 2\csc x\ \textup{Correct!} \end{gathered} math Click here to return to OFSA.
 * Question 12**
 * __Solution 12__**
 * __Error Explanation 12__**

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 * Question 13**
 * __Solution 13__**
 * __Error Explanation 13__**

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 * Question 14**
 * __Solution 14__**
 * __Error Explanation 14__**

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 * Question 15**
 * __Solution 15__**
 * __Error Explanation 15__**

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 * Question 16**
 * __Solution 16__**
 * __Error Explanation 16__**

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 * Question 17**
 * __Solution 17__**
 * __Error Explanation 17__**

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 * Question 18**
 * __Solution 18__**
 * __Error Explanation 18__**

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 * Question 19**
 * __Solution 19__**
 * __Error Explanation 19__**

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 * Question 20**
 * __Solution 20__**
 * __Error Explanation 20__**

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 * Question 21**
 * __Solution 21__**
 * __Error Explanation 21__**

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 * Question 22**
 * __Solution 22__**
 * __Error Explanation 22__**

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 * Question 23**
 * __Solution 23__**
 * __Error Explanation 23__**

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 * Question 24**
 * __Solution 24__**
 * __Error Explanation 24__**