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IBMaths: AA HLDPExam Questions3. Geometry & Trigonometry3.8 Further Trigonometry

Further Trigonometry (DP IB Maths: AA HL): Exam Questions

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1
Sme Calculator6 marks

Show that 

(i)
cos space left parenthesis theta plus straight pi over 2 right parenthesis equals negative sin space theta 

(ii)
tan left parenthesis theta minus pi right parenthesis equals tan space theta 

(iii)
sin space open parentheses theta minus straight pi over 4 close parentheses equals fraction numerator 1 over denominator square root of 2 end fraction left parenthesis sin space theta minus cos space theta right parenthesis

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2
Sme Calculator4 marks

Letspace f left parenthesis x right parenthesis equals tan space left parenthesis x plus pi right parenthesis space sin open parentheses x plus straight pi over 2 close parentheses where 0 less than x less than straight pi over 2.

By using the compound angle formulae, expressspace f left parenthesis x right parenthesis in terms of sin space x.

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3
Sme Calculator5 marks

Consider the equation cos space left parenthesis x minus 45 right parenthesis equals 2 space sin space x in the interval 0 less or equal than x less or equal than 360 degree. 

Find an exact value for tan space x.

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4a
Sme Calculator1 mark

Express cos space 4 theta in terms of cos space 2 theta.

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4b
Sme Calculator5 marks

Hence, show that cos space 4 theta equals 8 cos squared space theta space left parenthesis cos squared space theta minus 1 right parenthesis plus 1.

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5
Sme Calculator6 marks

Given that tan space A equals fraction numerator square root of 3 over denominator 2 end fraction comma solve the equation tan space left parenthesis A plus x right parenthesis equals 4 over 5  in the interval 0 less or equal than x less or equal than 360 degree.

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6
Sme Calculator6 marks

Prove that cos space 3 x identical to 4 space cos to the power of 3 space end exponent x minus 3 space cos space x.

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7
Sme Calculator7 marks

Solve the equation sin space 2 x minus cos space 2 x equals fraction numerator sin space x plus cos space x over denominator 2 end fraction minus 1 for the interval negative pi less than x less than 0.

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8a
Sme Calculator2 marks

Show that 1 minus cos space 2 x equals 2 minus 2 cos squared space x

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8b
Sme Calculator5 marks

Show that fraction numerator 1 over denominator cos space 2 x end fraction minus tan space 2 x equals fraction numerator cos space x minus sin space x over denominator cos space x plus sin space x end fraction

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9a
Sme Calculator3 marks

Find the exact values for tan space x given that tan squared space x plus 4 space tan space x plus 1 equals 0

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9b
Sme Calculator5 marks

Hence, solve the equation fraction numerator tan space x over denominator 2 space tan space x plus 1 end fraction equals tan space 2 x algebraically for the interval 0 less or equal than x less or equal than 2 pi.

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10a
Sme Calculator3 marks

The following diagram shows the triangle ABC where A B equals square root of 2, A C equals square root of 3 and straight B straight A with overparenthesis on top straight C equals 75 degree. q10a_3-8_medium_ib-maths-aa-hl

By writing 75 degree as 30 degree plus 45 degree find the value of sin left parenthesis 75 degree right parenthesis.

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10b
Sme Calculator4 marks

Find the area of the triangle, giving your answer in the form begin mathsize 16px style fraction numerator a plus square root of b over denominator c end fraction comma end style where a comma space b comma space c space element of space straight integer numbers.

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1
Sme Calculator7 marks

Solve the following equations in the given intervals. 

(i)
cos squared x minus sin squared x equals 1 half space space space space space space space space space space minus pi less or equal than x less or equal than pi
(ii)
fraction numerator 2 space tan space x over denominator 1 minus tan squared space x end fraction equals negative square root of 3 space space space space space space space space space space space space 0 less or equal than x less or equal than pi

           

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2a
Sme Calculator3 marks

Show that fraction numerator 5 space sin open parentheses pi plus x close parentheses over denominator tan space x end fraction identical to negative 5 space cos space x space space space space space space x not equal to fraction numerator n pi over denominator 2 end fraction.

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2b
Sme Calculator3 marks

Hence find the exact value of sin space xwhen fraction numerator 5 space sin open parentheses pi plus x close parentheses over denominator tan space x end fraction equals 4 for pi over 2 less or equal than x less or equal than pi.

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3a
Sme Calculator3 marks

Show that cos space x minus square root of 3 space sin space x identical to 2 space cos open parentheses x plus pi over 3 close parentheses.

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3b
Sme Calculator4 marks

Hence solve the equation 2 space cos open parentheses x plus pi over 3 close parentheses equals square root of 3 cos space x space sin space x minus 1 for the interval negative pi less or equal than x less or equal than pi. Show each step of your working.

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4
Sme Calculator5 marks

Show that 2 space cos space theta minus 5 space sin space theta can be written in the form R space cos open parentheses theta plus alpha close parentheses, where R greater than 0 and 0 less than alpha less than pi over 2.

Give R  in the form square root of k where k is an integer.

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5
Sme Calculator6 marks

Given that cos space x equals fraction numerator square root of 2 over denominator 5 end fraction and  pi less or equal than x less or equal than 2 pi, find the exact value of tan open parentheses 2 x close parentheses.

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6
Sme Calculator4 marks

Prove that 2 space sin space 2 theta open parentheses 2 space cos squared theta minus 1 close parentheses identical to sin space 4 theta.

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7
Sme Calculator7 marks

Solve the equation fraction numerator 1 over denominator cos squared space 2 x end fraction equals 1 plus tan space 2 x for the interval negative pi less than x less than 0.

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8a
Sme Calculator4 marks

Prove the identity fraction numerator 1 minus sin space 2 theta over denominator 2 space cos space 2 theta end fraction identical to fraction numerator 1 minus tan space theta over denominator 2 open parentheses 1 plus tan space theta close parentheses end fraction.

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8b
Sme Calculator3 marks

Solve the equation  fraction numerator 1 minus sin space 2 theta over denominator cos space 2 theta end fraction equals 2 square root of 2 for negative pi less or equal than theta less or equal than pi .

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9
Sme Calculator5 marks

Determine the value of the following expression

cos space 55 degree space cos space 5 degree minus cos space 85 degree space cos space 35 degree.

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10
Sme Calculator6 marks

The following diagram shows the triangle ABC where AC equals square root of 3 , straight A straight B with hat on top straight C equals 135 degree and straight B straight C with hat on top straight A equals 15 degree.

picture-2

Find the length AB giving your answer in the form fraction numerator a plus square root of b over denominator c end fraction, where a comma space b comma space c space element of space straight integer numbers.

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1
Sme Calculator5 marks

Find the value of arctan open parentheses 7 over 3 close parentheses minus arctan open parentheses 2 over 5 close parentheses.

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2
Sme Calculator6 marks

Solve 2 space tan space 2 x equals fraction numerator 2 tan space x over denominator 12 plus 5 space tan space x end fraction algebraically for negative 3 over 2 pi less than x less than pi over 2.

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3
Sme Calculator8 marks

By writing square root of 3 cos space theta plus sin space theta in the form R space sin open parentheses theta plus alpha close parentheses where R greater than 0 and 0 less than alpha less than pi over 2, or otherwise, solve 5 square root of 3 cos open parentheses negative straight pi over 12 close parentheses equals 2 x minus 5 space sin open parentheses negative straight pi over 12 close parentheses.

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4
Sme Calculator6 marks

Let f open parentheses x close parentheses equals tan open parentheses 2 x plus straight pi close parentheses sin open parentheses x minus straight pi close parentheses where 0 less than x less than straight pi over 2. 

Express f left parenthesis x right parenthesis in terms of sin space x .

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5
Sme Calculator8 marks

Given that  sin space A equals negative fraction numerator 1 over denominator square root of 3 end fraction for pi less or equal than A less or equal than fraction numerator 3 pi over denominator 2 end fractionand cos space B equals fraction numerator square root of 2 over denominator 3 end fraction for pi over 2 less or equal than B less or equal than pi , find the value of tan open parentheses A plus B close parentheses. Give the answer in the form square root of p open parentheses square root of q minus r close parentheses  where p, qand r are prime numbers.

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6a
Sme Calculator4 marks

By writing A equals fraction numerator A plus B over denominator 2 end fraction plus fraction numerator A minus B over denominator 2 end fraction and finding a similar expression for B, or otherwise, show that

sin space A plus sin space B equals 2 sin open parentheses fraction numerator A plus B over denominator 2 end fraction close parentheses cos open parentheses fraction numerator A minus B over denominator 2 end fraction close parentheses

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6b
Sme Calculator3 marks

Hence show that sin space fraction numerator 7 pi over denominator 12 end fraction plus sin space pi over 12 equals fraction numerator square root of 6 over denominator 2 end fraction.

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7
Sme Calculator8 marks

Prove that fraction numerator cos space 3 theta minus cos space theta over denominator sin space 3 theta minus sin space theta end fraction identical to negative tan space 2 theta.

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8a
Sme Calculator4 marks

By writing A equals fraction numerator A plus B over denominator 2 end fraction plus fraction numerator A minus B over denominator 2 end fraction and finding a similar expression for B, or otherwise, show that

cos space A minus cos space B equals negative 2 space sin open parentheses fraction numerator A plus B over denominator 2 end fraction close parentheses space sin open parentheses fraction numerator A minus B over denominator 2 end fraction close parentheses

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8b
Sme Calculator6 marks

Hence solve 4 space sin open parentheses fraction numerator 75 plus theta over denominator 2 end fraction close parentheses space sin open parentheses fraction numerator 75 minus theta over denominator 2 end fraction close parentheses equals 1 minus fraction numerator square root of 6 minus square root of 2 over denominator 2 end fraction for  negative 315 degree less or equal than theta less or equal than 45 degree.

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9a
Sme Calculator2 marks

The following diagram shows the triangle PQR where PQ equals 3, PR equals square root of 2 and straight R straight P with hat on top straight Q equals fraction numerator 5 pi over denominator 12 end fraction.

q9_ib-aa-hl_further-trigonometry_very_hard_diagram

Expand and simplify 1 half open parentheses 3 square root of 3 minus 1 close parentheses squared .

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9b
Sme Calculator8 marks

Show that the length of QR is fraction numerator 3 square root of 6 minus square root of 2 over denominator 2 end fractionunits.

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