OCR A Level: Core 3 - Trigonometry [January 2013 Q9]

Geometry Level 4

( i ) (\text{i}) Prove that cos ⁡ 2 ( θ + 45 ° ) − 1 2 ( cos ⁡ 2 θ − sin ⁡ 2 θ ) ≡ sin ⁡ 2 θ . \cos^2{(\theta + 45°)}-\dfrac{1}{2}(\cos 2 \theta - \sin 2 \theta) \equiv \sin^2 \theta .

( ii ) (\text{ii}) Hence solve 6 cos ⁡ 2 ( θ 2 + 45 ° ) − 3 ( cos ⁡ θ − sin ⁡ θ ) = 2 6\cos^2{ \left ( \dfrac{\theta}{2} + 45° \right )}-3(\cos \theta - \sin \theta) = 2 for − 90 ° < θ < 90 ° -90°< \theta < 90° .

( iii ) (\text{iii}) It is given that there are two values of θ \theta , where − 90 ° < θ < 90 ° -90°< \theta < 90° , satisfying 6 cos ⁡ 2 ( θ 3 + 45 ° ) − 3 ( cos ⁡ 2 θ 3 − sin ⁡ 2 θ 3 ) = k . 6\cos^2{\left (\frac{\theta}{3} + 45°\right )}-3\left (\cos \frac{2 \theta}{3} - \sin \dfrac{2 \theta}{3} \right ) = k.

Find the set of possible values for k k .

If a < k < b c a<k<\dfrac{b}{c} where a a , b b and c c are integers with b b and c c coprime, input a + b + c a+b+c as your answer.

There are 4 marks available for part (i), 3 marks for part (ii) and 3 marks for part (iii).
In total, this question is worth 13.9% of all available marks in the paper.
This is part of the set OCR A Level Problems .


The answer is 5.

Michael Fuller by Michael Fuller , 22, United Kingdom

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3 solutions

Michael Fuller
Mar 9, 2016

The mark scheme for this question: Large Version

1 Helpful 0 Interesting 0 Brilliant 0 Confused

( i ) C o s 2 ( θ + 45 ° ) − 1 2 ( C o s 2 θ − S i n 2 θ ) = ( S i n θ ∗ S i n 45 − C o s θ ∗ C o s 45 ) 2 − 1 2 ( C o s 2 θ − S i n 2 θ ) = ( 1 2 ∗ ( C o s θ − S i n θ ) 2 − 1 2 ( C o s 2 θ − S i n 2 θ ) = 1 2 ∗ ( 1 − S i n 2 θ ) − 1 2 ∗ C o s 2 θ + 1 2 ∗ S i n 2 θ ) = 1 2 − 1 2 ∗ ( 2 C o s 2 θ − 1 ) = 1 − C o s 2 θ ≡ S i n 2 θ \ \ (i)\\ Cos^2(\theta + 45°)-\dfrac1 2(Cos 2 \theta - Sin 2 \theta)\\ =(Sin \theta*Sin45-Cos\theta*Cos45)^2 - \dfrac1 2(Cos 2 \theta - Sin 2 \theta)\\ =(\dfrac 1 {\sqrt2}*(Cos\theta - Sin\theta)^2 - \dfrac1 2(Cos 2 \theta - Sin 2 \theta)\\ =\frac 1 2*(1 - \color{#3D99F6}{Sin2\theta)} - \frac 1 2 *Cos2\theta +\frac 1 2 *\color{#3D99F6}{Sin2\theta)} \\ = \frac 1 2 - \frac 1 2 *(2Cos^2\theta-1)=1- Cos^2\theta\\ \equiv Sin^2\theta\ \boxed{ } \\ ~~~\\ ( i i ) F r o m ( i ) a b o v e 6 cos ⁡ 2 ( θ 2 + 45 ° ) − 3 ( cos ⁡ θ − sin ⁡ θ ) = 2 ⟹ 6 S i n 2 θ 2 = 2 S i n θ 2 = ± 1 3 . ∴ θ = ± ( 2 ∗ 35.264 ) = 70.52 8 o . \ \ (ii)\\ From\ (i)\ above\ \ 6\cos^2{ \left ( \dfrac{\theta}{2} + 45° \right )}-3(\cos \theta - \sin \theta) = 2\\ \implies \ 6Sin^2\dfrac \theta 2=2\\ Sin\dfrac \theta 2=\pm \sqrt{\frac 1 3}.\\ \therefore\ \theta=\pm(2*35.264)=70.528^o.\\~~~~\\

( i i i ) − 90 ° < θ < 90 ° , ∴ − 3 0 o < θ 3 < 3 0 o . . . . . . . . . . . . . . . . ( ∗ ∗ ∗ ) F r o m ( i i ) a b o v e 6 C o s 2 ( θ 3 + 45 ° ) − 3 ( C o s 2 θ 3 − S i n 2 θ 3 ) = k ⟹ S i n 2 θ 3 = 1 6 ∗ k . ∴ S i n θ 3 = ± 1 6 ∗ k B u t f r o m ( ∗ ∗ ∗ ) S i n ( − 30 ) < θ 3 < S i n 30 ⟹ − 1 2 < ± θ 3 < 1 2 k c a n n o t b e n e g a t i v e ∴ 0 < k 6 < 1 2 ⟹ 0 < k < 6 4 = 2 3 . ∴ a + b + c = 0 + 2 + 3 = 5. \ \ (iii)\\ -90°< \theta < 90°, \ \ \therefore\ - 30^o<\dfrac \theta 3 < 30^o................(***)\\ From\ (ii)\ above\ \ 6Cos^2{\left (\frac{\theta}{3} + 45°\right )}-3\left (Cos \frac{2 \theta}{3} - Sin \frac{2 \theta}{3} \right ) = k\\ \implies\ Sin^2\dfrac \theta 3=\dfrac 1 6 *k.\\ \therefore\ Sin\dfrac \theta 3=\pm \sqrt{\dfrac 1 6 *k}\\ But\ from\ (***)\ Sin( - 30)<\dfrac \theta 3 < Sin30\ \ \implies\ \ - \frac 1 2<\pm\ \dfrac \theta 3 <\frac 1 2\\ k\ can\ not\ be\ negative\ \ \therefore 0<\sqrt{\dfrac k 6}<\frac 1 2\\ \implies\ 0<k<\dfrac 6 4=\frac 2 3.\\ \therefore\ a+b+c=0+2+3=5.

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Aditya Dhawan
Nov 2, 2016

The inequality between. K and 0 must be slack for 0 is very much an attainable minima, occurring at theta = 0

0 Helpful 0 Interesting 0 Brilliant 1 Confused

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