Trigonometri #7

7.       Soal-soal dan penyelesaiannya

1.        Misalkan $latex x$ adalah sudut dalam kuadran IV. Nyatakan semua perbandingan trigonometri dalam $latex {cosx}$.

Jawaban:

(1)     $latex {cosx=cosx}$

(2)   $latex secx=\frac{1}{cosx}$

(3)   $latex sin^2x=1-cos^2x$

$latex sinx=-\sqrt{1-cos^2x}$

(4)   $latex cscx=-\frac{1}{\sqrt{1-cos^2x}}=-\frac{\sqrt{1-cos^2x}}{1-cos^2x}=\frac{\sqrt{1-cos^2x}}{cos^2x-1}$

(5)   $latex tanx=\frac{sinx}{cosx}=-\frac{\sqrt{1-cos^2x}}{cosx}$

(6)   $latex cotx=\frac{cosx}{-\sqrt{1-cos^2x}}=\frac{cosx\sqrt{1-cos^2x}}{-(1-cos^2x)}=\frac{cosx\sqrt{1-cos^2x}}{cos^2x-1}$

2.      Buktikan $latex \frac{tan\alpha }{sec\alpha -1}=\frac{ sec\alpha +1}{tan\alpha }$.

Jawaban:

$latex \large\begin{array}{rcl}\frac{tan\alpha }{sec\alpha -1}&=&\frac{tan\alpha }{sec\alpha -1}\times \frac{sec\alpha +1}{sec\alpha +1}\\\\&=&\frac{tan\alpha (sec\alpha +1)}{sec^2\alpha -1}\\\\&=&\frac{tan\alpha (sec\alpha +1)}{tan^2\alpha }\\\\&=&\frac{ sec\alpha +1}{tan\alpha }\end{array}$

3.      Buktikan $latex \frac{sin^3\alpha +cos^3\alpha }{1-sin\alpha cos\alpha }= sin\alpha +cos\alpha$

Jawaban:

$latex \large\begin{array}{rcl}\frac{sin^3\alpha +cos^3\alpha }{1-sin\alpha cos\alpha }&=&\frac{sin^2\alpha .sin\alpha +cos^2\alpha .cos\alpha }{1-sin\alpha cos\alpha }\\\\&=&\frac{(1-cos^2\alpha )sin\alpha +(1-sin^2\alpha )cos\alpha }{1-sin\alpha cos\alpha }\\\\&=&\frac{sin\alpha -sin\alpha cos^2\alpha +cos\alpha -cos\alpha sin^2\alpha }{1-sin\alpha cos\alpha }\\\\&=&\frac{(sin\alpha +cos\alpha )-sin\alpha cos\alpha (cos\alpha +sin\alpha )}{1-sin\alpha cos\alpha }\\\\&=&\frac{(sin\alpha +cos\alpha )(1-sin\alpha cos\alpha )}{1-sin\alpha cos\alpha }\\\\&=&sin\alpha +cos\alpha \end{array}$

4.      Buktikan bahwa di dalam $latex \bigtriangleup ABC$.

1)       $latex sin\Big(\beta +\frac{1}{2}\alpha \Big)=sin\Big(\gamma +\frac{1}{2}\alpha \Big)$

2)     $latex cos\Big(\alpha +\frac{1}{2}\beta \Big)=-cos\Big(\gamma +\frac{1}{2}\beta \Big)$

Jawaban:

1)       Perhatikan $latex \bigtriangleup ABC$.

$latex \begin{array}{rcl}\beta +\frac{1}{2}\alpha +180^{\circ}-x&=&180^{\circ}\\\beta +\frac{1}{2}\alpha &=&x\qquad ..........\ (1)\end{array}$

$latex \begin{array}{rcl}\gamma +\frac{1}{2}\alpha +x&=&180^{\circ}\\x&=&180^{\circ}-(\gamma +\frac{1}{2}\alpha )\qquad ..........\ (2)\end{array}$

$latex \therefore \ sin\Big(\beta +\frac{1}{2}\alpha \Big)=sinx=sin\bigg(180^{\circ}-\Big(\gamma +\frac{1}{2}\alpha \Big)\bigg)=sin\Big(\gamma +\frac{1}{2}\alpha \Big)$

2)     Perhatikan $latex \bigtriangleup ABC$.

$latex \begin{array}{rcl}\alpha +\frac{1}{2}\beta +180^{\circ}-y&=&180^{\circ}\\\alpha +\frac{1}{2}\beta &=&y\qquad .......\ (1)\end{array}$

$latex \begin{array}{rcl}\gamma +\frac{1}{2}\beta +y&=&180^{\circ}\\y&=&180^{\circ}-\big(\alpha +\frac{1}{2}\beta \big)\qquad .......\ (2)\end{array}$

$latex \therefore \ cos\Big(\alpha +\frac{1}{2}\beta \Big)=cos\ y=cos\bigg(180^{\circ}-\Big(\gamma +\frac{1}{2}\beta \Big)\bigg)=-cos\Big(\gamma +\frac{1}{2}\beta \Big)$

5.      Diketahui $latex sin^2\beta =\frac{2q-7}{q+1}$. Tentukan batas-batas nilai $latex q$.

Jawaban:

$latex 0\leq sin^2\beta \leq 1$

$latex 0\leq \frac{2q-7}{q+1}\leq 1$

Bagian pertama:

$latex \frac{2q-7}{q+1}\geq 0\\\\q<-1\ atau\ q\geq \frac{7}{2}\qquad ........\ (1)$

Bagian kedua:

$latex \begin{array}{rcl}\frac{2q-7}{q+1}&\leq &1\\\\\frac{2q-7}{q+1}-1&\leq &0\\\\\frac{2q-7-(q+1)}{q+1}&\leq &0\\\\\frac{q-8}{q+1}&\leq &0\\\\-1< &q&\leq 8\qquad .......\ (2)\end{array}$

Batas-batas nilai $latex q$ yang memenuhi (1) dan (2) adalah $latex \frac{7}{2}\leq q\leq 8$.

Adjie Gumarang Pujakelana, 2013


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