STEP Support Programme. STEP 2 Trigonometry Questions: Solutions

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Angelina Lewis
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STEP Support Programme STEP Trigonometry Questions: Solutions We have cos cos cos sin sin cos cos sin cos sin cos cos cos sin cos cos cos cos cos cos cos + cos 4 cos

Similarly, using sin sin cos + cos sin leads to sin sin 4 sin i Using 4 sin sin sin gives us: α 7 sin sin α d 7 sin sin sin d α sin + sin d [ cos cos ] α [ cos α cos

$fractions with denominator 64 involved Another way is to show that 6c + is a factor of the equation you get by solving Iα Jα Remember that for a Show that you must$

1 STEP Support Programme STEP Trigonometry Questions: Solutions We have cos cos cos sin sin cos cos sin cos sin cos cos cos sin cos cos cos cos cos cos cos + cos 4 cos cos Since the answer is given, you do need to show every step Remember One equal sign per line, all equal signs aligned! Similarly, using sin sin cos + cos sin leads to sin sin 4 sin i Using 4 sin sin sin gives us: α 7 sin sin α d 7 sin sin sin d α sin + sin d [ cos cos ] α [ cos α cos α] [ ] c 4c c + 5 c + c + 5 Iα when c this can be easily found, as when α the integral is of the form f d giving cos ii If we call Eustace s attempt Jα we have: Jα α 7 sin sin d [ 7 sin sin 4 ] α 7 sin α sin 4 α 7 cos α cos α c 4 + c + You can then substitute c cosβ 6 and show that Iβ Jβ, but this is a little fiddly there are fractions with denominator 64 involved Another way is to show that 6c + is a factor of the equation you get by solving Iα Jα Remember that for a Show that you must fully support your answer STEP Trigonometry: Solutions

2 Equating Iα Jα gives: c + c + 5 c4 + c + c 4 6c c + 6c + * If you have already substituted c 6 into Iα and Jα and shown that this is a solution then you can factorise out 6c + without further eplanation of why you can do it Otherwise, you can: substitute c 6 into and show that this is a solution this is not too bad, especially if you don t evaluate 6 use long division to show that 6c + is a factor remember to show that the remainder is in this case! factorise out 6c + by inspection and then epand the brackets to show that this works In each case you do need to end by stating something like Hence c cosβ 6 gives Eustace the correct value of Iβ The equation fully factorises to give 6c+c+c, so we have cos α 6, cos α and cos α There are no given restrictions on α, so we need the general solutions These are: cos α α nπ cos α α π + nπ or α π + nπ cos α 6 α cos 6 + nπ or α cos 6 + nπ Another method would be to show that α c is a solution first and then factorise out c You could keep factorising until you had fully factorised and then show that c 6 is a solution See the topic notes STEP Trigonometry: Solutions

3 There are two basic approaches for the first identity Starting on the LHS gives: tan π 4 tan π 4 tan + tan π 4 tan tan + tan cos sin cos + sin Multiplying top and bottom by cos [ cos sin ] [ cos sin ] [ cos + sin ] [ cos sin ] sin cos cos sin sin cos sec tan Alternatively you can start on the RHS and use the t tan A formulae from the formula book with A : sec tan + t t t t t t + t t + t tan + tan tan π 4 i Setting π 4 into the identity from the stem gives: tan π 4 π 4 sec π 4 tan π 4 Then note that 4 π π + π which gives: tan π + π tan π + tan π tan π tan π + + as required 6 + ii Since the answer is given here, you can multiply across and verify that In this case verify means epand the brackets and show it is true You MUST show all the working here as it is a show that A table might be a nice clear way of showing your working STEP Trigonometry: Solutions

4 Alternatively you can rationalise the denominator : You should show a few more lines of working than presented here iii Here using 4π gives us tan 4π sec 4π tan 4π We know that tan 4 π and we also have sec θ + tan θ combining these gives us: tan 4 π You then need to carefully epand the squared bracket I would recommend a table for doing this to get to the required result If you had not been given the answer this is probably the approach you would have had to take The first approach probably has less room for error though! STEP Trigonometry: Solutions 4

5 Using the given substitution we have: a + a tan θ a sec θ dθ a dθ a θ + c a arctan a + c Since this result is in the stem of the question, you should epect to use it at least once and probably more often in the following question parts i a Using the substitution t sin gives us: π cos + sin d cos + t dt cos + t dt [ arctan t ] using the stem result π 4 b Using the suggested substitution which means that dt d sec gives us: t + 6t + t 4 dt tan + 6 tan + tan4 sec d tan + 6 tan + tan4 cos d cos sin cos cos + 6 sin + sin4 d cos sin cos sin cos + sin4 d cos sin cos + sin + 4 sin cos cos + sin d I d The last step came from using cos A cos A sin A, sin A sin A cos A and cos A + sin A STEP Trigonometry: Solutions 5

6 ii Using the substitution t tan in the same way as in part ib results in the integral cos + sin d A further substitution of y sin gives the integral: which gives the final answer as + y dy [ ] arctan y π 6 π 4 This is quite a long question and in my opinion quite a hard one as there is a lot going on i Using cos θ sin9 θ gives us sin9 θ sin 4θ There is one obvious solution ie 9 θ 4θ θ However the question asks for all the values of θ, so there are almost certainly some more Using the periodicity of sin we have: 9 θ 4θ + 6 θ 54 out of range 9 θ 4θ 6 θ 9 OK 9 θ 4θ 7 θ 6 OK 9 θ 4θ θ OK 9 θ 54 4θ θ 5 OK As is usually the case, a sketch helps make sure that you have all the possible values You can also do this by solving cos θ cos9 4θ and it might be a useful eercise to check that you get the same answers trying this method To find sin we use the double angle formulae to get: cos θ sin θ cos θ cos θ 4 sin θ cos θ sin θ Then either cos θ which is rejected as we are after θ or 4 sin θ sin θ It is usually helpful to write sin θ s so the equation is s 4s + To solve this cubic, first note that θ will be a solution from above, so s is a solution which we will reject as we want θ Using this to factorise we get: s 4s + s 4s + s Using the quadratic formula gives the other two solutions as: s ± ± 5 4 The final step is to notice that sin will be positive, so sin 4 5 STEP Trigonometry: Solutions 6

7 ii Writing the equation in terms of sin gives us: 4 sin + 4 sin cos 6 sin cos 6 sin sin Using sin s gives the quartic equation 6s 4 s + We can solve this for s giving s ± ± 9 4 ± 5 This looks a little problematic, but we do know that we want sin to be in the form p + q 5 You can solve p + q 5 ± 5 but, on the assumption that the parts of the question are related, it might be worth considering 4 5 Noting that means that we can deduce that the four values of 5 ± sin are ± 4 iii This is a hence question, so we need to refer to the previous parts and trying to do it another way will probably gain no credit If we take the equation from part ii and divide by 4 we get: sin + 4 sin If we then take α and 5α then this is the same as the equation in part iii 5α gives α 6 which gives which is a solution to part ii Finding the second solution is a little trickier If we are to use the equation in part ii then we need sin 5α ± We also want probably want to be related to A bit of trial and error is needed, but if we take + then we have sin 4 5, which is a solution to the equation in part ii This means α 9 66 which gives 5α and so this satisfies sin 5α Hence a second solution is α 66 STEP Trigonometry: Solutions 7

A Level Maths summer preparation work

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