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Trigonometry

Written on
November 2020 - Paper 2

Question 5

The graphs of $f(x)=-\frac{\displaystyle 1}{\displaystyle 2}\cos x$ and $g(x)=\sin (x+30^{\circ})$ , for the interval $x\in[0^{\circ}\,;\,180^{\circ}]$ , are drawn below. $A(130,9^{\circ}\,;\,0,33)$ is the approximate point of intersection of the two graphs.

$5.1$	Write down the period of $g$ .	$(1)$
$5.2$	Write down the amplitude of $f$ .	$(1)$
$5.3$	Determine the value of $f(180^{\circ})-g(180^{\circ})$	$(1)$
$5.4$	Use the graphs to determine the values of $x$ , in the interval $x\in[0^{\circ}\,;\,180^{\circ}]$ , for which:
$\quad 5.4.1$	$f(x-10^{\circ})=g(x-10^{\circ})$	$(1)$
$\quad 5.4.2$	$\sqrt{3}\sin x+\cos x \geq 1$	$(4)$
		$\textbf{[8]}$

Written on
November 2019 - Paper 2

Question 5

$5.1$	Simplify the following expression to ONE trigonometric term: $\frac{\displaystyle\sin{x}}{\displaystyle\cos{x}.\tan{x}}+\sin(180^{\circ}+x)\cos(90^{\circ}-x)$	$(5)$
$5.2$	$\textbf{Without using a calculator}$ , determine the value of: $\frac{\displaystyle \sin^2 35^{\circ}-\cos^2 35^{\circ}}{\displaystyle 4\sin 10^{\circ}\cos 10^{\circ}}$	$(4)$
$5.3$	Given: $\cos 26^{\circ}=m$ $\textbf{Without using a calculator}$ , determine $2\sin^2 77^{\circ}$ in terms of $m$ .	$(4)$
$5.4$	Consider: $f(x)=\sin(x+25^{\circ})\cos 15^{\circ}-\cos(x+25^{\circ})\sin 15^{\circ}$
$\quad 5.4.1$	Determine the general solution of $f(x)=\tan 165^{\circ}$	$(6)$
$\quad 5.4.2$	Determine the value(s) of $x$ in the interval $x\in[0^{\circ}\,;\,360^{\circ}]$ for which $f(x)$ will have a minimum value.	$(3)$
		$\textbf{[22]}$

Written on
November 2020 - Paper 2

Question 6

$6.1$	In the diagram, $P(-5\,;\,12)$ and $T$ lies on the positive $x$ -axis. $P\hat OT=\theta$

Answer the following without using a calculator:
$\quad 6.1.1$	Write down the value of $\tan\theta$	$(1)$
$\quad 6.1.2$	Calculate the value of $\cos\theta$	$(3)$
$\quad 6.1.3$	$S(a\,;\,b)$ is a point in the third quadrant such that $T\hat OS=\theta+90^{\circ}$ and $OS=6,5$ units. Calculate the value of $b$ .	$(4)$
$6.2$	Determine, without using a calculator, the value of the following trigonometric expression: $\frac{\displaystyle \sin{2x}.\cos(-x)+\cos{2x}.\sin(360^{\circ}-x)}{\displaystyle \sin(180^{\circ}+x)}$	$(5)$
$6.3$	Determine the general solution of the following equation: $6\sin^2 x+7\cos{x}-3=0$	$(6)$
$6.4$	Given: $x+\frac{\displaystyle 1}{\displaystyle x}=3\cos{A}$ and $x^2+\frac{\displaystyle 1}{\displaystyle x^2}=2$ Determine the value of $\cos{2A}$ without using a calculator.	$(5)$
		$\textbf{[24]}$

Written on
November 2019 - Paper 2

Question 6

In the diagram, the graphs of $f(x)=\sin x-1$ and $g(x)=\cos 2x$ are drawn for the interval $x\in[-90^{\circ}\,;\,360^{\circ}]$ . Graphs $f$ and $g$ intersect at $A$ . $B(360^{\circ}\,;\,-1)$ is a point on $f$ .

$6.1$	Write down the range of $f$ .	$(2)$
$6.2$	Write down the values of $x$ in the interval $x\in[-90^{\circ}\,;\,360^{\circ}]$ for which graph $f$ is decreasing.	$(2)$
$6.3$	$P$ and $Q$ are points on the graphs $g$ and $f$ respectively such that $PQ$ is parallel to the $y$ -axis. If $PQ$ lies between $A$ and $B$ , determine the value(s) of $x$ for which $PQ$ will be a maximum.	$(6)$
		$\textbf{[10]}$

Written on
November 2020 - Paper 2

Question 7

A landscape artist plans to plant flowers within two concentric circles around a vertical light pole $PQ$ . $R$ is a point on the inner circle and $S$ is a point on the outer circle. $R$ , $Q$ and $S$ lie in the same horizontal plane. $RS$ is a pipe used for the irrigation system in the garden. The radius of the inner circle is $r$ units and the radius of the outer circle is $QS$ . The angle of elevation from $S$ to $P$ is $30^{\circ}$ . $R\hat QS=2x$ and $PQ=\sqrt{3}r$

$7.1$	Show that $QS=3r$	$(3)$
$7.2$	Determine, in terms of $r$ , the area of the flower garden.	$(2)$
$7.3$	Show that $RS=r\sqrt{10-6\cos{2x}}$	$(3)$
$7.4$	If $r=10$ metres and $x=56^{\circ}$ , calculate $RS$ .	$(2)$
		$\textbf{[10]}$

Written on
November 2019 - Paper 2

Question 7

The diagram below shows a solar panel, $ABCD$ , which is fixed to a flat piece of concrete slab $EFCD$ . $ABCD$ and $EFCD$ are two identical rhombuses. $K$ is a point on $DC$ such that $DK=KC$ and $AK\perp DC$ . $AF$ and $KF$ are drawn. $A\hat DC=C\hat DE=60^{\circ}$ and $AD=x$ units.

$7.1$	Determine $AK$ in terms of $x$ .	$(2)$
$7.2$	Write down the size of $K\hat CF$ .	$(1)$
$7.3$	It is further given that $A\hat KF$ , the angle between the solar panel and the concrete slab, is $y$ . Determine the area of $\Delta AKF$ in terms of $x$ and $y$ .	$(7)$
		$\textbf{[10]}$

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