On the grid given in the ANSWER BOOK, draw the graph of \(g(x) = \cos 2x + 1\) for the interval \(x \in [-180^\circ\,;\,180^\circ]\). Clearly show the intercepts with the axes as well as the coordinates of the turning points.
About this question
Question 7.3.2 of Mathematics Grade 12 Paper 2, NSC Examination (DBE, November 2024). It tests the Term 1 ATP topic Trigonometry (Compound Angles, Identities, Equations).
- TermTerm 1
- ATP topicTrigonometry (Compound Angles, Identities, Equations)
- PaperPaper 2
- Marks3 marks
The graph of \(f(x) = \tan x\) is drawn for the interval \(x \in [-180^\circ\,;\,180^\circ]\). A second function is given in 7.3: \(g(x) = \cos 2x + 1\).
Diagram: the tan graph passes through \((-180^\circ\,;\,0)\), \((0^\circ\,;\,0)\) and \((180^\circ\,;\,0)\), with vertical dotted asymptotes at \(x = -90^\circ\) and \(x = 90^\circ\). The \(y\)-axis is marked from \(-3\) to \(3\).
The graph is \(\cos 2x\) shifted up by 1, so it ranges from 0 to 2.
Maximum turning points: \((-180^\circ\,;\,2)\), \((0^\circ\,;\,2)\) and \((180^\circ\,;\,2)\). The \(y\)-intercept is \((0\,;\,2)\).
Minimum turning points touch the \(x\)-axis: \((-90^\circ\,;\,0)\) and \((90^\circ\,;\,0)\). These are the only \(x\)-intercepts.
Smooth wave from \((-180^\circ\,;\,2)\) down to \((-90^\circ\,;\,0)\), up to \((0^\circ\,;\,2)\), down to \((90^\circ\,;\,0)\) and up to \((180^\circ\,;\,2)\).
More Questions On Trigonometry (Compound Angles, Identities, Equations)
\(\cos 35^\circ = p\), so draw a right triangle with hypotenuse 1 and adjacent side \(p\); the third side is \(\sqrt{1 - p^2}\), so:
\[\sin 35^\circ = \sqrt{1 - p^2}\]\[\sin 70^\circ = \sin 2(35^\circ) = 2\sin 35^\circ\cos 35^\circ\]\(\sin 70^\circ = 2p\sqrt{1 - p^2}\)
\(f\) is on or below the \(x\)-axis between its \(x\)-intercepts \(\pm 120^\circ\) and the ends of the interval.
\(-180^\circ \leq x \leq -120^\circ\) or \(120^\circ \leq x \leq 180^\circ\)
The graphs are equal at \(x\)-values \(-75^\circ\), \(45^\circ\) and \(165^\circ\). So \(x + 60^\circ\) is one of these.
For \(x \in [-45^\circ\,;\,0^\circ]\), \(x + 60^\circ \in [15^\circ\,;\,60^\circ]\), so:
\[x + 60^\circ = 45^\circ\]\(x = -15^\circ\)
In right-angled \(\triangle ABD\):
\[\sin\hat{B} = \frac{AD}{AB}\]\(AD = AB\sin\hat{B}\)
\(f'(x) \gt 0\) where \(f\) increases: \((-180^\circ\,;\,0^\circ)\). \(f'(x) \lt 0\) where \(f\) decreases: \((0^\circ\,;\,180^\circ)\).
\(g\) and \(f'\) must have opposite signs.
On \((-90^\circ\,;\,0^\circ)\): \(g \lt 0\) and \(f' \gt 0\).
On \((0^\circ\,;\,90^\circ)\): \(g \gt 0\) and \(f' \lt 0\).
\(-90^\circ \lt x \lt 0^\circ\) or \(0^\circ \lt x \lt 90^\circ\)
\(N(-75^\circ\,;\,k)\) lies on \(g\):
\[k = \sin 2(-75^\circ) = \sin(-150^\circ)\]\[= -\sin 30^\circ\]\(k = -\frac{1}{2}\)