Graph \(h\) is obtained when \(g\) is translated \(45^\circ\) to the right. Write down the equations of the asymptotes of \(h\) in the interval \(x \in [-180^\circ\,;\,180^\circ]\).
About this question
Question 7.6 of Mathematics Grade 12 Paper 2, NSC Examination (May/June) (DBE, May/June 2026). It tests the Term 1 ATP topic Trigonometry (Compound Angles, Identities, Equations).
- TermTerm 1
- ATP topicTrigonometry (Compound Angles, Identities, Equations)
- PaperPaper 2
- Marks2 marks
The graph of \(f(x) = 2\sin(x - 45^\circ)\) is drawn for \(x \in [-180^\circ\,;\,180^\circ]\). The \(y\)-axis is marked from \(-3\) to \(3\) and the \(x\)-axis at \(-180^\circ\), \(-90^\circ\), \(90^\circ\) and \(180^\circ\).
Diagram: \(f\) has a minimum value of \(-2\) at \(x = -45^\circ\) and a maximum value of \(2\) at \(x = 135^\circ\). It cuts the \(x\)-axis at \(x = -135^\circ\) and \(x = 45^\circ\), and the \(y\)-axis at \(-\sqrt{2}\).
The asymptotes of \(g\) at \(x = -90^\circ\) and \(x = 90^\circ\) move \(45^\circ\) to the right (\(x = -270^\circ\) moves to \(-225^\circ\), outside the interval).
\(x = -45^\circ\) and \(x = 135^\circ\)
More Questions On Trigonometry (Compound Angles, Identities, Equations)
\(\sin B\) is positive and \(90^\circ \lt B \lt 360^\circ\), so \(B\) is in the second quadrant: point \((x\,;\,4)\) with \(r = 5\).
\[x = -\sqrt{25 - 16} = -3, \text{ so } \cos B = -\frac{3}{5}\]\[\sin A = -\frac{4}{5} \text{ and } \cos A = -\frac{3}{5}\]\[\sin(A - B) = \sin A\cos B - \cos A\sin B\]\[= \left(-\frac{4}{5}\right)\left(-\frac{3}{5}\right) - \left(-\frac{3}{5}\right)\left(\frac{4}{5}\right)\]\[= \frac{12}{25} + \frac{12}{25}\]\(\sin(A - B) = \dfrac{24}{25}\)
\(= \frac{1}{2}(BC)(AB)\sin\hat{B}\)
\(\tan 50^\circ = \frac{k}{1}\): draw a right triangle with opposite side \(k\) and adjacent side 1.
Hypotenuse (Pythagoras):
\[r = \sqrt{k^2 + 1}\]\[\cos 40^\circ = \sin 50^\circ\]\(\cos 40^\circ = \dfrac{k}{\sqrt{k^2 + 1}}\)
\(\cos 35^\circ = p\), so draw a right triangle with hypotenuse 1 and adjacent side \(p\); the third side is \(\sqrt{1 - p^2}\), so:
\[\sin 35^\circ = \sqrt{1 - p^2}\]\[\sin 70^\circ = \sin 2(35^\circ) = 2\sin 35^\circ\cos 35^\circ\]\(\sin 70^\circ = 2p\sqrt{1 - p^2}\)
In \(\triangle ABD\):
\[\cos\alpha = \frac{BD}{AD} \Rightarrow AD = \frac{BD}{\cos\alpha}\]In \(\triangle ABC\):
\[\cos\alpha = \frac{BC}{AC} \Rightarrow AC = \frac{BC}{\cos\alpha}\]But \(BD = BC\) (given)
\(\therefore AD = AC\)