ONE EIGHTH CONTEST 2025
Ghana National College - 65 points
Tema Secondary School - 33 points
Effiduase SHCS - 28 points
ONE EIGHTH CONTEST 2025
Mankranso SHS - 43 points
St Paul's SHS, Denu - 23 points
Adonten SHS - 14 points
ONE EIGHTH CONTEST 2025
St. Thomas Aquinas SHS - 51 points
Tamale SHS - 37 points
O'Reilly SHS - 25 points
ROUND 2
TRIGONOMETRY
QUESTION
Given $\sin A=\dfrac{1}{2}$ and $\cos B=\dfrac{\sqrt2}{2}$ with $A$ and $B$ both acute, find $\cos(A-B)$.
ANSWER: $\dfrac{\sqrt6+\sqrt2}{4}$
SOLUTION 1
From the standard trigonometry table:
$\sin A=\dfrac{1}{2}$
$A=30^\circ$
$\cos B=\dfrac{\sqrt2}{2}$
$B=45^\circ$
Evaluate the expression:
$\cos(A-B)=\cos(30^\circ-45^\circ)$
$\cos(A-B)=\cos(-15^\circ)$
Since $\cos(-\theta)=\cos\theta$:
$\cos(A-B)=\cos15^\circ$
$\cos(A-B)=\dfrac{\sqrt6+\sqrt2}{4}$
SOLUTION 2
$\cos(A-B)=\cos A\cos B+\sin A\sin B$
Find the missing ratios using Pythagoras:
$\cos A=\dfrac{\sqrt{2^2-1^2}}{2}$
$\cos A=\dfrac{\sqrt3}{2}$
$\sin B=\dfrac{\sqrt{2^2-(\sqrt2)^2}}{2}$
$\sin B=\dfrac{\sqrt2}{2}$
Substitute:
$\cos(A-B)=\left(\dfrac{\sqrt3}{2}\times\dfrac{\sqrt2}{2}\right)+\left(\dfrac{1}{2}\times\dfrac{\sqrt2}{2}\right)$
$\cos(A-B)=\dfrac{\sqrt6}{4}+\dfrac{\sqrt2}{4}$
$\cos(A-B)=\dfrac{\sqrt6+\sqrt2}{4}$
NOTES (VALUES TO MEMORISE FOR SPEED ROUNDS)
$\cos15^\circ=\dfrac{\sqrt6+\sqrt2}{4}$
$\cos75^\circ=\dfrac{\sqrt6-\sqrt2}{4}$
$\cos105^\circ=\dfrac{\sqrt2-\sqrt6}{4}$
$\cos135^\circ=-\dfrac{\sqrt2}{2}$
$\cos165^\circ=-\dfrac{\sqrt6+\sqrt2}{4}$
$\sin15^\circ=\dfrac{\sqrt6-\sqrt2}{4}$
$\sin75^\circ=\dfrac{\sqrt6+\sqrt2}{4}$
$\sin105^\circ=\dfrac{\sqrt6+\sqrt2}{4}$
$\sin135^\circ=\dfrac{\sqrt2}{2}$
$\sin165^\circ=\dfrac{\sqrt6-\sqrt2}{4}$
$\tan15^\circ=2-\sqrt3$
$\tan75^\circ=2+\sqrt3$
$\tan105^\circ=-(2+\sqrt3)$
$\tan135^\circ=-1$
$\tan165^\circ=\sqrt3-2$
PRACTICE QUESTIONS
1. Given $\sin A=\dfrac{1}{2}$ and $\cos B=\dfrac{\sqrt2}{2}$ with $A=30^\circ$, $B=45^\circ$. Evaluate $\cos(A-B)$.
ANSWER: $\dfrac{\sqrt6+\sqrt2}{4}$
SOLUTION
$\cos(A-B)=\cos A\cos B+\sin A\sin B$
$=\dfrac{\sqrt3}{2}\times\dfrac{\sqrt2}{2}+\dfrac{1}{2}\times\dfrac{\sqrt2}{2}$
$\cos(30^\circ-45^\circ)=\cos(15^\circ)=\dfrac{\sqrt6+\sqrt2}{4}$
2. Given $\sin A=\dfrac{\sqrt2}{2}$ and $\cos B=\dfrac{1}{2}$ with $A=45^\circ$, $B=60^\circ$. Evaluate $\cos(A-B)$.
ANSWER: $\dfrac{\sqrt6+\sqrt2}{4}$
SOLUTION
$\cos(A-B)=\cos A\cos B+\sin A\sin B$
$=\dfrac{\sqrt2}{2}\times\dfrac{1}{2}+\dfrac{\sqrt2}{2}\times\dfrac{\sqrt3}{2}$
$\cos(45^\circ-60^\circ)=\cos(15^\circ)=\dfrac{\sqrt6+\sqrt2}{4}$
3. Given $\sin A=\dfrac{1}{2}$ and $\cos B=\dfrac{1}{2}$ with $A=30^\circ$, $B=60^\circ$. Evaluate $\cos(A+B)$.
ANSWER: $0$
SOLUTION
$\cos(A+B)=\cos A\cos B-\sin A\sin B$
$=\dfrac{\sqrt3}{2}\times\dfrac{1}{2}-\dfrac{1}{2}\times\dfrac{\sqrt3}{2}$
$\cos(30^\circ+60^\circ)=\cos(90^\circ)=0$
4. Given $\sin A=\dfrac{\sqrt2}{2}$ and $\cos B=\dfrac{\sqrt3}{2}$ with $A=45^\circ$, $B=30^\circ$. Evaluate $\cos(A+B)$.
ANSWER: $\dfrac{\sqrt6-\sqrt2}{4}$
SOLUTION
$\cos(A+B)=\cos A\cos B-\sin A\sin B$
$=\dfrac{\sqrt2}{2}\times\dfrac{\sqrt3}{2}-\dfrac{\sqrt2}{2}\times\dfrac{1}{2}$
$\cos(45^\circ+30^\circ)=\cos(75^\circ)=\dfrac{\sqrt6-\sqrt2}{4}$
5. Given $\sin A=\dfrac{\sqrt3}{2}$ and $\cos B=\dfrac{\sqrt2}{2}$ with $A=60^\circ$, $B=45^\circ$. Evaluate $\cos(A-B)$.
ANSWER: $\dfrac{\sqrt6+\sqrt2}{4}$
SOLUTION
$\cos(A-B)=\cos A\cos B+\sin A\sin B$
$=\dfrac{1}{2}\times\dfrac{\sqrt2}{2}+\dfrac{\sqrt3}{2}\times\dfrac{\sqrt2}{2}$
$\cos(60^\circ-45^\circ)=\cos(15^\circ)=\dfrac{\sqrt6+\sqrt2}{4}$
6. Given $\sin A=\dfrac{\sqrt3}{2}$ and $\cos B=\dfrac{\sqrt2}{2}$ with $A=60^\circ$, $B=45^\circ$. Evaluate $\cos(A+B)$.
ANSWER: $-\dfrac{\sqrt6-\sqrt2}{4}$
SOLUTION
$\cos(A+B)=\cos A\cos B-\sin A\sin B$
$=\dfrac{1}{2}\times\dfrac{\sqrt2}{2}-\dfrac{\sqrt3}{2}\times\dfrac{\sqrt2}{2}$
$\cos(60^\circ+45^\circ)=\cos(105^\circ)=-\dfrac{\sqrt6-\sqrt2}{4}$