# Trigonometry (Practice Problems)

1.        Given that $\displaystyle \sin 16{}^\circ =p,$ determine the following in terms of $\displaystyle p$.

(a) $\displaystyle \sin 196{}^\circ$.

(b) $\displaystyle \cos 16{}^\circ$.

(c) $\displaystyle \tan 32{}^\circ$.

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 $\displaystyle \begin{array}{l}\ \ \ \ \ \ \ \sin 16{}^\circ =p\\\\\text{(a)}\ \ \ \sin 196{}^\circ =\sin (180{}^\circ +16{}^\circ )\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\sin 16{}^\circ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-p\\\\\\\text{(b)}\ \ \ \cos 16{}^\circ \ \ =\sqrt{{1-{{{\sin }}^{2}}16{}^\circ }}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sqrt{{1-{{p}^{2}}}}\\\\\\\text{(c)}\ \ \ \tan 16{}^\circ \ \ =\displaystyle \frac{{\sin 16{}^\circ }}{{\cos 16{}^\circ }}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{p}{{\sqrt{{1-{{p}^{2}}}}}}\\\\\ \ \ \ \ \ \ \tan 32{}^\circ \ \ =\displaystyle \frac{{2\tan 16{}^\circ }}{{1-{{{\tan }}^{2}}16{}^\circ }}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{\displaystyle \frac{{2p}}{{\sqrt{{1-{{p}^{2}}}}}}}}{{1-\displaystyle \frac{{{{p}^{2}}}}{{1-{{p}^{2}}}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{2p}}{{\sqrt{{1-{{p}^{2}}}}}}\times \displaystyle \frac{{1-{{p}^{2}}}}{{1-2{{p}^{2}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{2p\sqrt{{1-{{p}^{2}}}}}}{{1-2{{p}^{2}}}}\end{array}$

2.        Simplify $\displaystyle \frac{{\sqrt{{1-{{{\cos }}^{2}}2\theta }}}}{{\cos (-\theta )\cos (90{}^\circ +\theta )}}$ completely, given that $\displaystyle 0{}^\circ <\theta <90{}^\circ$.

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 $\displaystyle \begin{array}{l}\ \ \ \ \ \ 1-{{\cos }^{2}}2\theta ={{\sin }^{2}}2\theta \\\\\ \ \ \ \ \ \cos (-\theta )=\cos \theta \\\\\ \ \ \ \ \ \cos (90{}^\circ +\theta )=-\sin \theta \\\\\therefore \ \ \ \ \ \displaystyle \frac{{\sqrt{{1-{{{\cos }}^{2}}2\theta }}}}{{\cos (-\theta )\cos (90{}^\circ +\theta )}}\\\\\ \ \ \ \ \ \ \ =\displaystyle \frac{{\sqrt{{{{{\sin }}^{2}}2\theta }}}}{{\cos \theta \left( {-\sin \theta } \right)}}\\\\\ \ \ \ \ \ \ \ =-\displaystyle \frac{{\sin 2\theta }}{{\cos \theta \sin \theta }}\\\\\ \ \ \ \ \ \ \ =-\displaystyle \frac{{2\sin \theta \cos \theta }}{{\cos \theta \sin \theta }}\\\\\ \ \ \ \ \ \ \ =-2\end{array}$

3.        In the diagram $\displaystyle \vartriangle TPS,\ PR\bot TS$. If $\displaystyle TR=2,RS = 3$ and $\displaystyle PR=1$, find the value of

(a) $\displaystyle \sin T$ and $\displaystyle \cos T$

(b) $\displaystyle \sin S$ and $\displaystyle \cos S$

Hence without using tables, calculate the value of $\displaystyle \cos (S+T)$.

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 $\displaystyle \begin{array}{l}\ \ \ \ \ \vartriangle TPS,\ PR\bot TS\\\\\ \ \ \ \ TR=2,RS=3,\ PR=1\\\\\therefore \ \ \ PT=\sqrt{{{{2}^{2}}+{{1}^{2}}}}=\sqrt{5}\\\\\ \ \ \ \ PS=\sqrt{{{{3}^{2}}+{{1}^{2}}}}=\sqrt{{10}}\\\\\therefore \ \ \ \sin T=\displaystyle \frac{1}{{\sqrt{5}}},\ \cos T=\displaystyle \frac{2}{{\sqrt{5}}}\\\\\ \ \ \ \ \sin S=\displaystyle \frac{1}{{\sqrt{{10}}}},\ \cos S=\displaystyle \frac{3}{{\sqrt{{10}}}}\\\\\therefore \ \ \ \cos \left( {S+T} \right)=\cos S\cos T-\sin S\sin T\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{3}{{\sqrt{{10}}}}\times \displaystyle \frac{2}{{\sqrt{5}}}-\displaystyle \frac{1}{{\sqrt{{10}}}}\times \displaystyle \frac{1}{{\sqrt{5}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{5}{{\sqrt{{50}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{5}{{5\sqrt{2}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{1}{{\sqrt{2}}}\end{array}$

4.        If $\displaystyle \sin x-\cos x=\frac{3}{4}$, calculate the value of $\displaystyle \sin 2x$ without using tables.

Show/Hide Solution
 $\displaystyle \begin{array}{l}\ \ \ \ \sin x-\cos x=\displaystyle \frac{3}{4}\\\\\therefore \ \ {{\left( {\sin x-\cos x} \right)}^{2}}=\displaystyle \frac{9}{{16}}\\\\\ \ \ \ {{\sin }^{2}}x-2\sin x\cos x+{{\cos }^{2}}x=\displaystyle \frac{9}{{16}}\\\\\ \ \ \ {{\sin }^{2}}x+{{\cos }^{2}}x-2\sin x\cos x=\displaystyle \frac{9}{{16}}\\\\\therefore \ \ 1-\sin 2x=\displaystyle \frac{9}{{16}}\\\\\therefore \ \ \sin 2x=\displaystyle \frac{7}{{16}}\end{array}$

5.        Simplify $\displaystyle \frac{{4\sin x\cos x}}{{2{{{\sin }}^{2}}x-1}}$ to a single trigonometric ratio. Hence, find the value of $\displaystyle \frac{{4\sin 15{}^\circ \cos 15{}^\circ }}{{2{{{\sin }}^{2}}15{}^\circ -1}}$ in surd form.

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 $\displaystyle \begin{array}{l}\ \ \ \ \displaystyle \frac{{4\sin x\cos x}}{{2{{{\sin }}^{2}}x-1}}\\\\=\ \ \displaystyle \frac{{2\left( {2\sin x\cos x} \right)}}{{-\left( {1-2{{{\sin }}^{2}}x} \right)}}\\\\=\ \ \displaystyle \frac{{2\sin 2x}}{{-\cos 2x}}\\\\=\ \ -2\tan 2x\\\\\therefore \ \ \displaystyle \frac{{4\sin 15{}^\circ \cos 15{}^\circ }}{{2{{{\sin }}^{2}}15{}^\circ -1}}\\\\=\ -2\tan 30{}^\circ \\\\=\ -\displaystyle \frac{2}{{\sqrt{3}}}\end{array}$

6.        Points $\displaystyle P(-\sqrt{7},3)$ and $\displaystyle S(a,b)$ are on the Cartesian plane, as shown in figure. If $\displaystyle \angle POR=\angle POS=\theta$, $\displaystyle \angle POR=\angle POS=\theta$ and $\displaystyle OS=6$,

(a) $\displaystyle \tan \theta$

(b) $\displaystyle \sin (-\theta )$

(c) $\displaystyle a$

Show/Hide Solution
 $\displaystyle \begin{array}{l}\ \ \ \ \ \ P=(-\sqrt{7},3)\\\\\therefore \ \ \ \ OP=\sqrt{{7+9}}=4\\\\\therefore \ \ \ \ \sin \theta =\displaystyle \frac{3}{4}\\\\\text{(a)}\ \ \tan \theta =-\displaystyle \frac{3}{{\sqrt{7}}}\\\\\text{(b)}\ \ \sin \left( {-\theta } \right)=-\sin \theta =-\displaystyle \frac{3}{4}\\\\\text{(c)}\ \ \cos 2\theta =1-2{{\sin }^{2}}\theta =1-\displaystyle \frac{9}{8}=-\displaystyle \frac{1}{8}\\\\\ \ \ \ \ \text{Since}\ \cos 2\theta =\displaystyle \frac{a}{6},\ \\\\\ \ \ \ \ \displaystyle \frac{a}{6}=-\displaystyle \frac{1}{8}\\\\\therefore \ \ \ a\ =-\displaystyle \frac{3}{4}\end{array}$

7.        If $\displaystyle \sin A = p$ and $\displaystyle \cos A = q,$

(a) Show that $\displaystyle \tan 2A=\frac{{2pq}}{{{{q}^{2}}-{{p}^{2}}}}$

(b) Express $\displaystyle p^4-q^4$ as a single trigonometric ratio.

Show/Hide Solution
 $\displaystyle \begin{array}{l}\ \ \ \ \ \ \sin A=p,\ \ \cos A=q\\\\\therefore \ \ \ \ \tan A=\displaystyle \frac{p}{q}\ \\\\\text{(a)}\ \ \tan 2A=\ \ \displaystyle \frac{{2\tan A}}{{1-{{{\tan }}^{2}}A}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \ \displaystyle \frac{{2\left( {\displaystyle \frac{p}{q}} \right)}}{{1-{{{\left( {\displaystyle \frac{p}{q}} \right)}}^{2}}}}\ \ \\\ \ \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \ \displaystyle \frac{{\displaystyle \frac{{2p}}{q}}}{{\displaystyle \frac{{{{q}^{2}}-{{p}^{2}}}}{{{{q}^{2}}}}}}\ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \ \displaystyle \frac{{2p}}{q}\times \displaystyle \frac{{{{q}^{2}}}}{{{{q}^{2}}-{{p}^{2}}}}\ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \ \displaystyle \frac{{2pq}}{{{{q}^{2}}-{{p}^{2}}}}\\\\\text{(b)}\ \ {{p}^{4}}-{{q}^{4}}=\left( {{{p}^{2}}+{{q}^{2}}} \right)\left( {{{p}^{2}}-{{q}^{2}}} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left( {{{{\sin }}^{2}}A+{{{\cos }}^{2}}A} \right)\left( {{{{\sin }}^{2}}A-{{{\cos }}^{2}}A} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-1\left( {{{{\cos }}^{2}}A-{{{\sin }}^{2}}A} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\cos 2A\end{array}$

8.       In the diagram below, $\displaystyle T(x,p)$ is a point in the third quadrant and it is given that $\displaystyle \sin \alpha =\frac{p}{{\sqrt{{1+{{p}^{2}}}}}}$.

(a) Show that $\displaystyle x =-1$.

(b) Find $\displaystyle \cos 2\alpha$ in terms of $\displaystyle p$.

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 $\displaystyle \begin{array}{l}\text{(a)}\ \ \sin \alpha =\displaystyle \frac{p}{{\sqrt{{1+{{p}^{2}}}}}}\ \ \ \left[ {\because \text{given}} \right]\\\\\ \ \ \ \ \sin \alpha =\displaystyle \frac{p}{h}\ \ \ \left[ {\text{diagram}} \right]\\\\\therefore \,\ \ \displaystyle \frac{p}{h}=\displaystyle \frac{p}{{\sqrt{{1+{{p}^{2}}}}}}\\\\\therefore \ \ h=\sqrt{{1+{{p}^{2}}}}\\\\\ \ \ \ \text{Since}\ {{x}^{2}}+{{p}^{2}}={{h}^{2}},\\\\\ \ \ \ {{x}^{2}}+{{p}^{2}}=1+{{p}^{2}}\\\\\therefore \ \ {{x}^{2}}=1\\\\\therefore \ \ x=\pm 1\\\\\ \ \ \ \text{Since}\ \alpha \,\ \text{lies in }{{\text{3}}^{{\text{rd}}}}\text{ quadrant,}\\\\\ \ \ x=-1.\\\\\\\text{(b)}\ \ \cos 2\alpha =1-2{{\sin }^{2}}\alpha \\\\\therefore \,\ \ \ \cos 2\alpha =1-\displaystyle \frac{{2{{p}^{2}}}}{{1+{{p}^{2}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{1-{{p}^{2}}}}{{1+{{p}^{2}}}}\end{array}$

9.        If $\displaystyle \sin 76{}^\circ = x$ and $\displaystyle \cos 76{}^\circ = y$, show that $\displaystyle x^2- y^2 = \sin 62{}^\circ$.

Show/Hide Solution
 $\displaystyle \begin{array}{l}\ \ \ \ \sin 76{}^\circ =x,\ \cos 76{}^\circ =y\\\\\ \ \ {{x}^{2}}-{{y}^{2}}={{\sin }^{2}}76{}^\circ -{{\cos }^{2}}76{}^\circ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\left( {{{{\cos }}^{2}}76{}^\circ -{{{\sin }}^{2}}76{}^\circ } \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\cos \left( {2\times 76{}^\circ } \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\cos 152{}^\circ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\cos \left( {90{}^\circ +62{}^\circ } \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\left( {-\sin 62{}^\circ } \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sin 62{}^\circ \end{array}$

10.      $\displaystyle T(8, k)$ is a point in the first quadrant as shown in figure. If $\displaystyle \tan β = \frac{1}{4},$ determine, without using tables,

(a) the value of $\displaystyle k$.

(b) the value of $\displaystyle \sin β$ in surd form.

Show/Hide Solution
 $\displaystyle \begin{array}{l}\,\ \ \tan \beta =\displaystyle \frac{1}{4}\ \ \left[ {\text{given}} \right]\\\\\,\ \ \tan \beta =\displaystyle \frac{k}{8}\ \ \left[ {\text{diagram}} \right]\\\\\therefore \displaystyle \frac{k}{8}=\displaystyle \frac{1}{4}\\\\\therefore \ k=2\\\\\ \ \ \text{Since}\ O{{T}^{2}}={{8}^{2}}+{{k}^{2}},\\\\\ \ \ O{{T}^{2}}={{8}^{2}}+{{2}^{2}}=68\\\\\therefore \ OT=\sqrt{{68}}=2\sqrt{{17}}\\\\\therefore \ \sin \beta =\displaystyle \frac{k}{{OT}}\\\\\ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{2}{{2\sqrt{{17}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{1}{{\sqrt{{17}}}}\end{array}$