IBDP Maths SL 3.6 Pythagorean identities AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Part (a)
Show \( \cos \hat{A} – \sin \hat{A} = \frac{1}{\sqrt{2}} \).
In \( \triangle ABC \), \( \angle ABC = 90^\circ \), \( AC = \sqrt{2} \), \( AB = BC + 1 \).
Let \( BC = x \), so \( AB = x + 1 \).
\[ \cos \hat{A} = \frac{AB}{AC} = \frac{x + 1}{\sqrt{2}}, \quad \sin \hat{A} = \frac{BC}{AC} = \frac{x}{\sqrt{2}} \]
\[ \cos \hat{A} – \sin \hat{A} = \frac{x + 1}{\sqrt{2}} – \frac{x}{\sqrt{2}} = \frac{1}{\sqrt{2}} \]
Answer: \( \cos \hat{A} – \sin \hat{A} = \frac{1}{\sqrt{2}} \).
Part (b)
Square both sides of \( \cos \hat{A} – \sin \hat{A} = \frac{1}{\sqrt{2}} \):
\[ \cos^2 \hat{A} – 2 \cos \hat{A} \sin \hat{A} + \sin^2 \hat{A} = \frac{1}{2} \]
Since \( \cos^2 \hat{A} + \sin^2 \hat{A} = 1 \):
\[ 1 – 2 \cos \hat{A} \sin \hat{A} = \frac{1}{2} \implies \sin \hat{A} \cos \hat{A} = \frac{1}{4} \]
Using \( \sin 2\hat{A} = 2 \sin \hat{A} \cos \hat{A} \):
\[ \sin 2\hat{A} = \frac{1}{2} \]
For \( 0^\circ < \hat{A} < 90^\circ \), solutions are:
\[ 2\hat{A} = 30^\circ \implies \hat{A} = 15^\circ \quad \text{or} \quad 2\hat{A} = 150^\circ \implies \hat{A} = 75^\circ \]
Since \( \angle ABC = 90^\circ \), \( \angle A + \angle C = 90^\circ \).
If \( \hat{A} = 15^\circ \), then \( \angle C = 75^\circ \). If \( \hat{A} = 75^\circ \), then \( \angle C = 15^\circ \).
Answer: Angles are \( 15^\circ \), \( 75^\circ \), \( 90^\circ \).
Part (c)
Apply Pythagoras’ theorem: \( AC^2 = AB^2 + BC^2 \).
Let \( BC = x \), so \( AB = x + 1 \), \( AC = \sqrt{2} \):
\[ 2 = (x + 1)^2 + x^2 = 2x^2 + 2x + 1 \]
\[ 2x^2 + 2x – 1 = 0 \]
\[ x = \frac{-2 \pm \sqrt{12}}{4} = \frac{-1 \pm \sqrt{3}}{2} \]
Since \( BC > 0 \), \( x = \frac{\sqrt{3} – 1}{2} \).
\[ \sin \hat{A} = \frac{BC}{AC} = \frac{\frac{\sqrt{3} – 1}{2}}{\sqrt{2}} = \frac{\sqrt{6} – \sqrt{2}}{4} \]
Answer: \( BC = \frac{\sqrt{3} – 1}{2} \), \( \sin \hat{A} = \frac{\sqrt{6} – \sqrt{2}}{4} \).
Part (d)
Find perpendicular from \( B \) to \( AC \).
Method 1: \( h = AB \sin \hat{A} \).
\[ AB = \frac{\sqrt{3} + 1}{2}, \quad \sin \hat{A} = \frac{\sqrt{6} – \sqrt{2}}{4} \]
\[ h = \frac{\sqrt{3} + 1}{2} \cdot \frac{\sqrt{6} – \sqrt{2}}{4} = \frac{\sqrt{18} – \sqrt{6} + \sqrt{6} – \sqrt{2}}{8} = \frac{\sqrt{2}}{4} \]
Method 2: Area method: \( \frac{1}{2} \cdot AB \cdot BC = \frac{1}{2} \cdot AC \cdot h \).
Area: \( \frac{1}{2} \cdot \frac{\sqrt{3} + 1}{2} \cdot \frac{\sqrt{3} – 1}{2} = \frac{1}{4} \).
\[ \frac{1}{2} \cdot \sqrt{2} \cdot h = \frac{1}{4} \implies h = \frac{\sqrt{2}}{4} \]
Answer: Perpendicular length is \( \frac{\sqrt{2}}{4} \).
Question
Show that \(\frac{{\cos A + \sin A}}{{\cos A – \sin A}} = \sec 2A + \tan 2A\) .
▶️Answer/Explanation
Markscheme
METHOD 1
\(\frac{{\cos A + \sin A}}{{\cos A – \sin A}} = \sec 2A + \tan 2A\)
consider right hand side
\(\sec 2A + \tan 2A = \frac{1}{{\cos 2A}} + \frac{{\sin 2A}}{{\cos 2A}}\) M1A1
\( = \frac{{{{\cos }^2}A + 2\sin A\cos A + {{\sin }^2}A}}{{{{\cos }^2}A – {{\sin }^2}A}}\) A1A1
Note: Award A1 for recognizing the need for single angles and A1 for recognizing \({\cos ^2}A + {\sin ^2}A = 1\) .
\( = \frac{{{{(\cos A + \sin A)}^2}}}{{(\cos A + \sin A)(\cos A – \sin A)}}\) M1A1
\( = \frac{{\cos A + \sin A}}{{\cos A – \sin A}}\) AG
METHOD 2
\(\frac{{\cos A + \sin A}}{{\cos A – \sin A}} = \frac{{{{(\cos A + \sin A)}^2}}}{{(\cos A + \sin A)(\cos A – \sin A)}}\) M1A1
\( = \frac{{{{\cos }^2}A + 2\sin A\cos A + {{\sin }^2}A}}{{{{\cos }^2}A – {{\sin }^2}A}}\) A1A1
Note: Award A1 for correct numerator and A1 for correct denominator.
\( = \frac{{1 + \sin 2A}}{{\cos 2A}}\) M1A1
\( = \sec 2A + \tan 2A\) AG
[6 marks]
Question
a. Show that \(\cot \alpha = \tan \left( {\frac{\pi }{2} – \alpha } \right)\) for \(0 < \alpha < \frac{\pi }{2}\). [1]
b. Hence find \(\int_{\tan \alpha }^{\cot \alpha } {\frac{1}{{1 + {x^2}}}{\text{d}}x,{\text{ }}0 < \alpha < \frac{\pi }{2}} \). [4]
▶️Answer/Explanation
Markscheme
EITHER
a.use of a diagram and trig ratios
eg,
\(\tan \alpha = \frac{O}{A} \Rightarrow \cot \alpha = \frac{A}{O}\)
from diagram, \(\tan \left( {\frac{\pi }{2} – \alpha } \right) = \frac{A}{O}\) R1
OR
use of \(\tan \left( {\frac{\pi }{2} – \alpha } \right) = \frac{{\sin \left( {\frac{\pi }{2} – \alpha } \right)}}{{\cos \left( {\frac{\pi }{2} – \alpha } \right)}} = \frac{{\cos \alpha }}{{\sin \alpha }}\) R1
THEN
\(\cot \alpha = \tan \left( {\frac{\pi }{2} – \alpha } \right)\) AG
[1 mark]
\(\int_{\tan \alpha }^{\cot \alpha } {\frac{1}{{1 + {x^2}}}{\text{d}}x} = [\arctan x]_{\tan \alpha }^{\cot \alpha }\) (A1)
Note: Limits (or absence of such) may be ignored at this stage.
\( = \arctan (\cot \alpha ) – \arctan (\tan \alpha )\) (M1)
\( = \frac{\pi }{2} – \alpha – \alpha \) (A1)
\( = \frac{\pi }{2} – 2\alpha \) A1
[4 marks]