P5.8 Trig proofs

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question was a good differentiator at the later end of the paper. Success with this question was somewhat mixed, with some learners scoring most marks in part (b), whilst part (a) proved to be more challenging as it tested not only their knowledge of trigonometric identities but also their algebraic manipulation skills. In part (a), most responses attempted the proof by using both compound angle expansions to reach an expression in sin x and cos x, usually gaining the method mark and in most cases also the first accuracy mark. Once a correct expression was achieved, some stronger responses reached the given result by simplifying the terms. The most common mistakes committed while implementing this approach, were sign errors, especially for the compound angle expansion of ( ) cos 30 x + ° , incorrect multiplication by 3 and bracketing errors, which were not condoned in this ‘show’ question. A smaller proportion of responses used the R-alpha method with ( ) cos 30 R x + °± alpha , which required learners to find both R and alpha correctly to reach ( ) 2cos 30 30 2cos x x + °- ° = as requested.

Common mistakes

• expand both compound angles (sign of cos(x + 30°) expansion), use sin 30° = ½ and cos 30° = √3/2 exactly
• bracket errors not condoned
• R-alpha method needs R = 2 and the angle

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

A significant number of candidates found this question challenging. In part (a) the first challenge candidates faced was combining the two terms to produce a single fraction. This was completed successfully by many candidates, either as a fraction in cosec or a fraction in sin . The second challenge was using trigonometric identities correctly. The majority used 2 2 sin cos 1 + = but the more successful candidates typically used 2 2 1 cot cosec + = . It was only a minority who managed to complete the proof successfully. It was possible to complete the proof starting either from either side of the identity, however, generally those candidates who started with the left-hand side were more successful. Few, if any, candidates gave a completely correct solution starting from the right-hand side and the majority only obtained the mark for using a correct Pythagorean identity. A small number of candidates used a “meet in the middle” approach with varying degrees of success. Notational errors were not widespread, but variables were sometimes missing, or indices misplaced resulting in the loss of the final mark. Many also lost the accuracy mark because they did not show all the necessary steps by proceeding to the given answer using sine and cosine.

Common mistakes

• combine into one fraction 2cosecθ/(cosec²θ − 1)
• use cosec²θ − 1 = cot²θ (or sin² + cos² = 1)
• convert to sin and cos to reach 2tanθsecθ
• no notational slips (cos²θ not cosθ²)
• starting from the RHS rarely succeeded

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

A significant number of candidates found this question challenging. In part (b), candidates that spotted the connection with part (a) tended to achieve at least 3 out of the 4 marks available. Those who failed to get the first line right with the variable of 2x on both sides but just substituted 2tan sec x x or even 2tan sec on the left-hand side gained no marks in this part. Candidates took a number of different approaches to solving the correct equation. The most frequent was to obtain 2 1 tan 2 2 x = but approaches involving 2 3sin 2 1 x = or 2 3cos 2 2 x = were relatively common. There were a few candidates who used the identity 2 2tan tan 2 1 tan x x x = - to reach a quadratic in 2 tan .x These attempts were usually successful. Some candidates stated that the square root of 1 2 is 1 4 which lost them the final two marks in this part. However, those candidates that did get to one of the correct equations could normally proceed to one correct answer (which was typically x = 17.6°). It was only a minority of candidates who gave both correct answers. The final mark was regularly lost because only one solution was identified, or candidates had incorrectly found 45 degrees by incorrectly solving sec2x = 0 and interpreting this as cos2x = 0 rather than being undefined. Most candidates that achieved marks in this part of the questions could correctly identify the order of operations to solve their equations although there were some candidates who struggled with this. For the final mark, a few wrote down 17.7 instead of 17.6 and some had extra answers in the given range and therefore lost the accuracy mark. A small minority of candidates did not show appropriate working in part (b), sometimes merely writing down the two correct answers to the given equation. The question explicitly asked for all stages of a candidate’s working as the answers can be found using a calculator. Hence candidates that gave the correct answers without appropriate working were not awarded any marks in this part of the question.

Common mistakes

• replace θ by 2x: 2tan2x sec2x = cot2x sec2x
• sec2x ≠ 0 (not 45° from cos2x = 0)
• tan²2x = ½ (or sin²2x = ⅓)
• x = 17.6°, 72.4° (both, 1 d.p., none extra)
• answers without working score 0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

A number of candidates lost all marks on this question by not attempting any solution. Most candidates correctly expanded the given expression and used sin2 x + cos2 x = 1 to simplify and score the first two marks. A small number incorrectly used (sin x - cos x)2 = sin2 x - cos2 x, scoring no marks. Others attempted to substitute values of 90 and 180 into the given expression instead of using algebraic manipulation. Whilst most candidates scored the first two marks, only the very best were able to score the final mark, with most candidates failing to offer a convincing reason for a contradiction. Candidates who drew a graph of sin x and cos x or sin 2x to support their understanding tended to score well. Successful candidates typically explained that 2x must be between 180 and 360, and hence sin 2x must be negative, which is a contradiction with their inequality. Those that worked with sin x cos x were generally more successful in achieving this mark than those who chose to use sin2x. However, it was clear that not all students understood what an obtuse angle was. Another typical correct conclusion involved candidates stating if x is obtuse, sin x must be positive and cos x must be negative, leading to a negative product and a contradiction. Of those who were able to offer a reason for the contradiction, some still lost the final A mark for failing to then reach a conclusion that the original statement was true or because of errors in the proof such as missing x’s or mixed variables.

Common mistakes

• expand (sin x − cos x)² = sin²x − 2sinxcosx + cos²x (not sin²x − cos²x)
• use sin²x + cos²x = 1 to get −2sinxcosx < 0
• obtuse so sin x > 0, cos x < 0 (or 180° < 2x < 360°, sin2x < 0): contradiction, so sin x − cos x ≥ 1
• substituting 90° and 180° is not a proof
• many did not explain the contradiction

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This was found to be another discriminating question. Those who were familiar with the topic made good progress yet there were many other attempts that scored no marks at all. Part(a) There were a significant number of fully correct solutions for this part, although some candidates did not achieve the stage where both the numerator and denominators of their fraction were factorised correctly with a common factor of ( sin theta +cos theta ). Another difficulty encountered by some, was how to use the different forms of the identity for cos2 theta in order to deal with, and remove, the presence of both “1’s ” in the fraction. Very few notational errors were seen, which was a pleasing improvement.

Common mistakes

• choose cos 2θ = 1 − 2sin²θ on top and 2cos²θ − 1 below to remove the 1s
• factorise to cancel (sin θ + cos θ)
• no notation errors (cosθ², mixed variables)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This was found to be another discriminating question. Those who were familiar with the topic made good progress yet there were many other attempts that scored no marks at all. Part (b) A significant number of candidates used the result of (a) correctly and started their solution with tan2x=3sin2x. However, it was also the case that tanx =3 sin2x was seen several times as well. Of those candidates who started correctly with tan2x = 3sin2x, the significant majority simply cancelled out the resulting factor of sin2x and in so doing neglected to consider the possibility that sin2x=0 and in so doing the solution x = 90° was omitted from the fully correct list of three answers. Many candidates however, did state that cos2x =⅓ and went on to get 35.3° and 144.7° and so only one mark was lost.

Common mistakes

• replace θ by 2x: tan 2x = 3 sin 2x (not tan x)
• factorise sin 2x(1 − 3cos 2x) = 0, do not cancel sin 2x (loses x = 90°)
• x = 35.3°, 90°, 144.7°

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (a) was generally answered well and there were remarkably few solutions containing missing variables. On this occasion, candidates who started with the RHS fared better as those who worked from the LHS occasionally missed the required step between 𝑐𝑐𝑐𝑐𝑐𝑐2𝜃𝜃/sin𝜃𝜃 and cos𝜃𝜃cot𝜃𝜃, while those who worked from both sides would often get confused or, if they progressed correctly they simply failed to make the necessary conclusion.

Common mistakes

• cosec θ = 1/sin θ, single fraction (1 − sin²θ)/sin θ = cos²θ/sin θ, then cos θ × cos θ/sin θ
• show the step to cos θ cot θ
• working on both sides needs a conclusion

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Many students were able to get started in part (a) with correct use of cos(2A + A) alongside the addition formula. Candidates who managed this, regularly went on to apply the double angle formulae correctly to achieve an expression in cosA only. Often circuitous routes were taken, most notably with the use of cos2A = cos2A - sin2A and then making a second substitution for sin2A (this sometimes resulted in bracketing and/or sign errors). A small number of candidates added a proof for the double angle formulae from the addition formulae even though this was unnecessary. A high proportion of students who managed to get the first mark went on to get at least 3 if not 4 marks in this part of the question. It was very much an all or nothing question with many only getting the first mark. A significant number presented their work in a way that was sometimes difficult to follow even if they did manage to score full marks.

Common mistakes

• write cos 3A = cos(2A + A) and use the compound angle formula
• cos 2A = 2cos²A − 1 and sin²A = 1 − cos²A
• rigorous with no mixed variables or missing brackets

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Q12(a) discriminated well between the higher ability candidates and it was possible for candidates of all abilities to gain access to Q12(b). In Q12(a), many candidates struggled to apply a complete strategy which would help them to make significant progress in proceeding from cos3 sin3 , via an sin cos 2 2 cos2 (cos sin ) intermediate stage of cos(3 ) , cos2 , or to the given sin cos sin cos sin cos result 2cot2 . Candidates who used Way 1, as described in the mark scheme, gained some credit for rationalising cos3 sin3 to give cos3 cos sin3 sin . Of those candidates sin cos sin cos 1 sin cos sin 2 who correctly applied to the denominator, only a few realised that 2 A common error in using Way 1 was to the numerator could be written as cos2 . 2 2 cos 3 sin 3 simplify cos3 sin3 or cos4 sin 4 . to give sin cos sin cos sin cos Way 2, as described in the mark scheme (and variations of this), was the most popular approach among candidates who completed the proof successfully. Candidates only started to access the marks for Way 2 when they combined their fractions by rationalising the denominator. Errors using Way 2 included the incorrect expansion of cos(2 ) or sin(2 ) , bracketing errors, manipulation errors or sign errors. They usually progressed as far as A few candidates started their proof from 2cot2 . 2 2(1 tan ) 2cos2 2cot 2 , 2cot 2 writing either or but could make no sin 2 2tan creditable progress until they applied a Way 3 method as described in the mark scheme. Some candidates spent a considerable amount of time attempting to write 3 3 4cos 3cos 3sin 4sin cos3 sin3 in terms of to give e.g. . sin cos sin cos candidates only started to access the marks for this method when they combined their fractions by rationalising the denominator. In Q13(b), most candidates used the result given in Q13(a) to deduce the equation 1 2cot 2 4. tan 2 , Most of these then attempted to solve and some obtained the 2 Common errors included attempting to solve tan2 correct answer 103.3 . 2 A few candidates obtained by . and finding more than one value for 103.3 2 2(1 tan ) 4. solving the equation 2tan

Common mistakes

• common denominator sinθcosθ
• numerator cos3θcosθ + sin3θsinθ = cos(3θ − θ) = cos2θ
• sinθcosθ = ½sin2θ
• cos3θ + sin3θ wrongly as cos4θ ...

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question discriminated well between students of all abilities with part (a) more accessible to students than part (b). In part (a), most students provided a proof which progressed from one side of the identity to the other by using methodical steps and correct trigonometric identities. There was no preference seen for starting their proof on a particular side of the identity. Some students manipulated both sides of the identity to give a common expression 2 2sin , with many providing some indication that their proof was complete.

Common mistakes

• tan θ = sin θ / cos θ and sin 2θ = 2 sin θ cos θ
• cos 2θ = 1 − 2sin²θ
• show all steps to 2sin²θ

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question discriminated well between students of all abilities with part (a) more accessible to students than part (b). Part (b) proved to be discriminating, with some students not realising that they could use the result from part (a). Some students substituted 1 cos2x with tan sin2 x x and then proceeded to cancel tan sin2 x xfrom both sides, while other students substituted tan sin2 x x with 1 cos2x and then proceeded to cancel 1 cos2x from both sides. These two strategies together with applying the identity 2 2 sec 1 tan x x usually led to a 3-term quadratic in tan .x Most students used a correct method, usually by factorising, to solve their quadratic equation in tan x but only a few stated all three correct solutions. As a result of cancelling either tan sin2 x x or 1 cos2x from both sides of their equation, many students missed out the correct solution 0. x Those students who factorised out these terms were usually more successful in finding 0. x Some students did not give their other two solutions in the required form of the exact 4 x and 1.326 x to 3 decimal places.

Common mistakes

• replace (1 − cos 2x) by tan x sin 2x (or vice versa)
• factorise, do not cancel, so that x = 0 is kept
• sec²x = 1 + tan²x gives tan²x − 3tan x − 4 = 0
• x = −π/4 (exact) and 1.326