Skip to content

Course home

Sign up

3 Unit A2 1: Pure Mathematics (A-level only)

EasyMediumHard
123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150151152153154155156157158159160161162163164165166167168169170171172173174175176177178179180181182183184185186187188189190191192193194195196197198199200201202203204205206207208209210211212213214215216217218219220221222223224225226227228229230231232233234235236237238239240241242243244245246247248249250251252253254255256257258259260261262263264265266267268269270271272273274275276277278279280281282283284285286287288289290291292293294295296297298299300301302303304305306307308309310311312313314315316317318319
Question 75

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

a.

Two acoustic waves are superimposed such that their resulting phase angle ϕ\phiϕ satisfies the equation

6sin⁡(ϕ−60∘)=2cos⁡(ϕ+45∘) \sqrt{6} \sin(\phi - 60^\circ) = 2 \cos(\phi + 45^\circ) 6​sin(ϕ−60∘)=2cos(ϕ+45∘)

Show that

tan⁡ϕ=52+3 \tan \phi = \frac{5}{2 + \sqrt{3}} tanϕ=2+3​5​

and hence that

tan⁡ϕ=10−53 \tan \phi = 10 - 5\sqrt{3} tanϕ=10−53​
[5]
b.

Hence or otherwise, solve for 0≤θ<180∘0 \le \theta < 180^\circ0≤θ<180∘,

6sin⁡(3θ−60∘)=2cos⁡(3θ+45∘) \sqrt{6} \sin(3\theta - 60^\circ) = 2 \cos(3\theta + 45^\circ) 6​sin(3θ−60∘)=2cos(3θ+45∘)

giving your answers to one decimal place.

[4]
Markscheme

3 Unit A2 1: Pure Mathematics (A-level only) Questions

  1. A Level
  2. /Maths
  3. /3 Unit A2 1: Pure Mathematics (A-level only)

Question bank