Skip to content

Course home

Sign up

1 Unit 1: Pure Mathematics A

EasyMediumHard
123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150151152153154155156157158159160161162163164165166167168169170171172173174175176177178179180181182183184185186187188189190191192193194195196197198199200201202203204205206207208209210211212213214215216217218219220221222223224225226227228229230231232233234235236237238239240241242243244245246247248249250251252253254255256257258259260261262263264265266267268269270271272273274275276277278279280281282283284285286287288289290291292293294295296297298299300301302303304305306307308309310311312313314315316317318319320321322323324325326327328329330331332333334335336337338339340341342343344345346347348349350351352353354355356357358359360361362363364365366367368369370371372373374375376377378379380381382383384385386387388389390391392393394395396397398399400401402403404405406407408409410411412413414415416
Question 225
i.

A mechanical sensor measures the angular misalignment α \alpha\,α of a rotating component. The misalignment follows the equilibrium condition:

6sin⁡(α−35∘)=−5cos⁡(α−35∘) 6 \sin(\alpha - 35^\circ) = -5 \cos(\alpha - 35^\circ) 6sin(α−35∘)=−5cos(α−35∘)

Determine all possible values of α \alpha\,α in the range 0<α<360∘0 < \alpha < 360^\circ0<α<360∘, giving your answers to one decimal place.

[4]
ii a.

Show that the equation describing the lateral vibration of a specific piston,

5sin⁡3x=7sin⁡x−4sin⁡xcos⁡x 5 \sin^3 x = 7 \sin x - 4 \sin x \cos x 5sin3x=7sinx−4sinxcosx

can be written in the form

sin⁡x(acos⁡2x+bcos⁡x+c)=0 \sin x (a \cos^2 x + b \cos x + c) = 0 sinx(acos2x+bcosx+c)=0

where aaa, b b\,b and c c\,c are integers to be found.

[3]
ii b.

Hence find all real solutions for −π≤x≤π-\pi \le x \le \pi−π≤x≤π to the equation

5sin⁡3x=7sin⁡x−4sin⁡xcos⁡x 5 \sin^3 x = 7 \sin x - 4 \sin x \cos x 5sin3x=7sinx−4sinxcosx

giving your answers to two decimal places where appropriate.

[3]
Markscheme

1 Unit 1: Pure Mathematics A Questions

  1. A Level
  2. /Maths
  3. /1 Unit 1: Pure Mathematics A

Question bank