Trigonometri #6

6.       Batasan nilai perbandingan trigonometri

Batas-batas nilai perbandingan trigonometri terdefinisi sebagai berikut:

$latex {\color{Blue} \begin{array}{rcl}-1\leqslant sinx\leqslant 1\qquad &\Rightarrow& \qquad cscx \geq 1\ atau\ cscx< -1\\\\-1\leqslant cosx\leqslant 1\qquad &\Rightarrow& \qquad secx \geq 1\ atau\ secx< -1\\\\-\infty \leqslant tanx\leqslant +\infty \qquad &\Rightarrow& \qquad -\infty\leq cotx \leq +\infty\end{array}}$

Contoh-contoh soal

1.        Diketahi $latex sinx=\frac{p-2}{8-p}$. Tentukan batas-batas nilai $latex p$.

Jawaban:

$latex sinx=\frac{p-2}{8-p}\qquad \Rightarrow \qquad -1\leq \frac{p-2}{8-p}\leq 1$

Bagian pertama:

$latex \large{\begin{array}{rcl}\frac{p-2}{8-p}&\leq &1 \\\\\frac{p-2}{8-p}-1&\leq &0\\\\\frac{p-2-(8-p)}{8-p}&\leq &0\\\\\frac{2p-10}{8-p}&\leq &0\\\\\frac{2(p-5)}{8-p}&\leq &0\\\\p\leq 5&atau&p> 8 \end{array}}$

Bagian kedua:

$latex \large{\begin{array}{rcl}\frac{p-2}{8-p}&\geq &-1 \\\\\frac{p-2}{8-p}+1&\geq &0\\\\\frac{p-2}{p-8}-1&\leq &0\\\\\frac{p-2-(p-8)}{p-8}&\leq &0\\\\\frac{6}{p-8}&\leq &0\\\\p&\leq &8 \end{array}}$

Karena $latex p> 8$ dan $latex p< 8$ saling menggugurkan maka batasan nilai $latex p$ yang memenuhi adalah $latex p\leq 5$.

2.      Untuk sudut $latex \varphi$ dalam kuadran III diketahui $latex tan\varphi =\frac{1}{2}\sqrt{2}$. Carilah perbandingan trigonometri lainnya dari $latex \varphi$.

Jawaban:

(1)     $latex tan\varphi =\frac{1}{2}\sqrt{2}$

(2)   $latex cot\varphi =\frac{2}{\sqrt{2}}=\sqrt{2}$

(3)   $latex sec^2\varphi =1+tan^2\varphi =1+\frac{1}{4}.2=\frac{6}{4}$

$latex sec\varphi =-\sqrt{\frac{6}{4}}=-\frac{1}{2}\sqrt{6}$

(4)   $latex cos\varphi =-\frac{2}{\sqrt{6}}=-\frac{1}{3}\sqrt{6}$

(5)   $latex cos^2\varphi +sin^2\varphi =1$

$latex \frac{6}{9}+sin^2\varphi =1$

$latex sin^2\varphi =1-\frac{6}{9}=\frac{3}{9}$

$latex sin\varphi =-\sqrt{\frac{3}{9}}=-\frac{1}{3}\sqrt{3}$

(6)   $latex csc\varphi =-\frac{3}{\sqrt{3}}=-\sqrt{3}$

3.      Untuk sudut $latex \delta $ dalam kuadran IV, nyatakan semua perbandingan trigonometri dalam $latex cot\delta $.

Jawaban:

(1)     $latex cot\delta = cot\delta $

(2)   $latex tan\delta = \frac{1}{cot\delta }$

(3)   $latex csc^2\delta = cot^2\delta +1$

$latex csc\delta = -\sqrt{cot^2\delta +1}$

(4)   $latex sin\delta = -\frac{1}{\sqrt{cot^2\delta +1}}= -\frac{\sqrt{cot^2\delta +1}}{cot^2\delta +1}$

(5)   $latex cos^2\delta +sin^2\delta =1$

$latex cos^2\delta +\frac{1}{cot^2\delta +1}=1$

$latex cos^2\delta +\frac{1}{cot^2\delta +1}=1$

$latex cos\delta =-\frac{cot\delta }{\sqrt{cot^2\delta +1}}=-\frac{cot\delta \sqrt{cot^2\delta +1}}{cot^2\delta +1}$

(6)     $latex sec\delta =-\frac{\sqrt{cot^2\delta +1}}{cot\delta }$

Adjie Gumarang Pujakelana, 2013


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