\(\text{cosec }\theta= \dfrac{1}{\text{sin }\theta}\)
\(\text{sec }\theta= \dfrac{1}{\text{cos }\theta}\)
\(\text{cot }\theta= \dfrac{1}{\text{tan }\theta}\)
The angles \(A\) and \(B\) are complementary angles to each other if \(A+B=90^\circ\). Hence,
\(A=90^\circ-B\) and \(B=90^\circ-A\)
The diagram shows the reference angles, \(\alpha\) for the angles \( 0° \leqslant \theta \leqslant 360^{\circ}\) or \( 0° \leqslant \theta \leqslant 2\pi\).
The trigonometric ratios of special angles \(30^\circ\), \(45^\circ\) and \(60^\circ\) can be determined by using right-angled triangles.
(a)
\(\begin{aligned} \text{cos }\theta&= \text{sin } (90^{\circ}-\theta)\\ \text{cos }13^{\circ}&= \text{sin } (90^{\circ}-13^{\circ})\\ &= \text{sin }77^{\circ}\\ &= 0.9744. \end{aligned}\)
(b)
\(\begin{aligned} \text{cosec }\theta&= \text{sec } (90^{\circ}-\theta)\\ \text{cosec }27^{\circ}&= \text{sec } (90^{\circ}-27^{\circ})\\ &= \text{sec }63^{\circ}\\ &= \dfrac{1}{\text{cos }63^{\circ}}\\ &= \dfrac{1}{k}. \end{aligned}\)
Use the unit circle and state the value of \(\text{cos }135^{\circ}\).
The coordinates that correspond to \(135^{\circ}\) are
\(\begin{pmatrix} -\dfrac{1}{\sqrt2}, \ \dfrac{1}{\sqrt2} \end{pmatrix}\) and \(\text{cos }135^{\circ} = x\text{-coordinate}\).
Hence, \(\text{cos }135^{\circ} = -\dfrac{1}{\sqrt2}\).
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