IB Mathematics AA Pythagorean identities Study Notes
IB Mathematics AA Pythagorean identities Study Notes
IB Mathematics AA Pythagorean identities Study Notes Offer a clear explanation of Pythagorean identities , including various formula, rules, exam style questions as example to explain the topics. Worked Out examples and common problem types provided here will be sufficient to cover for topic Pythagorean identities.
Pythagorean Identities
The Pythagorean identities are fundamental relationships in trigonometry, derived from the geometry of the unit circle. This section focuses on these identities, double angle identities, and their applications.
Key Concepts
- Pythagorean Identity:
- \(\cos^2\theta + \sin^2\theta = 1\).
- Double Angle Identities:
- \(\sin(2\theta) = 2\sin\theta\cos\theta\).
- \(\cos(2\theta) = \cos^2\theta – \sin^2\theta = 2\cos^2\theta – 1 = 1 – 2\sin^2\theta\).
- Relationships Between Trigonometric Ratios:
- Use algebraic manipulation and Pythagorean identities to find relationships between \(\sin\theta\), \(\cos\theta\), and \(\tan\theta\) without explicitly calculating \(\theta\).
Guidance, Clarifications, and Syllabus Links
- Illustrations: Use geometrical diagrams and dynamic graphing software to visualize the relationships between trigonometric functions and identities.
- Applications:
- Verify identities using algebraic techniques.
- Derive relationships between trigonometric functions for specific cases.
- Focus: Solve problems using identities, especially for simplifying expressions and verifying equations.
Examples
Example 1: Using the Pythagorean Identity
- Given \(\sin\theta = \frac{3}{5}\), find \(\cos\theta\):
- \(\cos^2\theta = 1 – \sin^2\theta = 1 – \left(\frac{3}{5}\right)^2 = 1 – \frac{9}{25} = \frac{16}{25}\).
- \(\cos\theta = \pm\frac{4}{5}\) (choose the sign based on the quadrant).
Example 2: Double Angle Identity
- Given \(\cos x = \frac{3}{4}\) and \(x\) is acute, find \(\sin(2x)\):
- \(\sin x = \sqrt{1 – \cos^2x} = \sqrt{1 – \left(\frac{3}{4}\right)^2} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}\).
- \(\sin(2x) = 2\sin x \cos x = 2 \cdot \frac{\sqrt{7}}{4} \cdot \frac{3}{4} = \frac{3\sqrt{7}}{8}\).
IB Mathematics AA SL Pythagorean Identities Exam Style Worked Out Questions
Question
In the triangle ABC, \({\rm{A\hat BC}} = 90^\circ\) , \({\text{AC}} = \sqrt {\text{2}}\) and AB = BC + 1.
a.Show that cos \(\hat A – \sin \hat A = \frac{1}{{\sqrt 2 }}\). [3]
b.By squaring both sides of the equation in part (a), solve the equation to find the angles in the triangle. [8]
c.Apply Pythagoras’ theorem in the triangle ABC to find BC, and hence show that \(\sin \hat A = \frac{{\sqrt 6 – \sqrt 2 }}{4}\). [6]
d. Hence, or otherwise, calculate the length of the perpendicular from B to [AC]. [4]
▶️Answer/Explanation
Markscheme
a.\(\cos \hat A = \frac{{{\text{BA}}}}{{\sqrt 2 }}\) A1
\(\sin \hat A = \frac{{{\text{BC}}}}{{\sqrt 2 }}\) A1
\(\cos \hat A – \sin \hat A = \frac{{{\text{BA}} – {\text{BC}}}}{{\sqrt 2 }}\) R1
\( = \frac{1}{{\sqrt 2 }}\) AG
[3 marks]
\({\cos ^2}\hat A – 2\cos \hat A\sin \hat A + {\sin ^2}\hat A = \frac{1}{2}\) M1A1
\(1 – 2\sin \hat A\cos \hat A = \frac{1}{2}\) M1A1
\(\sin 2\hat A = \frac{1}{2}\) M1
\(2\hat A = 30^\circ \) A1
angles in the triangle are 15° and 75° A1A1
Note: Accept answers in radians.
[8 marks]
\({\text{B}}{{\text{C}}^2} + {({\text{BC}} + 1)^2} = 2\) M1A1
\(2{\text{B}}{{\text{C}}^2} + 2{\text{BC}} – 1 = 0\) A1
\({\text{BC}} = \frac{{ -2 + \sqrt {12} }}{4}\left( { = \frac{{\sqrt 3 – 1}}{2}} \right)\) M1A1
\(\sin \hat A = \frac{{{\text{BC}}}}{{\sqrt 2 }} = \frac{{\sqrt 3 – 1}}{{2\sqrt 2 }}\) A1
\( = \frac{{\sqrt 6 – \sqrt 2 }}{4}\) AG
[6 marks]
EITHER
\(h = {\text{ABsin}}\hat A\) M1
\( = ({\text{BC}} + 1)\sin \hat A\) A1
\( = \frac{{\sqrt 3 + 1}}{2} \times \frac{{\sqrt 6 – \sqrt 2 }}{4} = \frac{{\sqrt 2 }}{4}\) M1A1
OR
\(\tfrac{1}{2}AB.BC = \tfrac{1}{2}AC.h\) M1
\(\frac{{\sqrt 3 – 1}}{2} \cdot \frac{{\sqrt {3 + 1} }}{2} = \sqrt {2h} \) A1
\(\frac{2}{4} = \sqrt 2 h\) M1
\(h = \frac{1}{{2\sqrt 2 }}\) A1
[4 marks]
Question
In the triangle ABC, \({\rm{A\hat BC}} = 90^\circ\) , \({\text{AC}} = \sqrt {\text{2}}\) and AB = BC + 1.
a.Show that cos \(\hat A – \sin \hat A = \frac{1}{{\sqrt 2 }}\). [3]
b.By squaring both sides of the equation in part (a), solve the equation to find the angles in the triangle. [8]
c.Apply Pythagoras’ theorem in the triangle ABC to find BC, and hence show that \(\sin \hat A = \frac{{\sqrt 6 – \sqrt 2 }}{4}\). [6]
d.Hence, or otherwise, calculate the length of the perpendicular from B to [AC]. [4]
▶️Answer/Explanation
Markscheme
a. \(\cos \hat A = \frac{{{\text{BA}}}}{{\sqrt 2 }}\) A1
\(\sin \hat A = \frac{{{\text{BC}}}}{{\sqrt 2 }}\) A1
\(\cos \hat A – \sin \hat A = \frac{{{\text{BA}} – {\text{BC}}}}{{\sqrt 2 }}\) R1
\( = \frac{1}{{\sqrt 2 }}\) AG
[3 marks]
\({\cos ^2}\hat A – 2\cos \hat A\sin \hat A + {\sin ^2}\hat A = \frac{1}{2}\) M1A1
\(1 – 2\sin \hat A\cos \hat A = \frac{1}{2}\) M1A1
\(\sin 2\hat A = \frac{1}{2}\) M1
\(2\hat A = 30^\circ \) A1
angles in the triangle are 15° and 75° A1A1
Note: Accept answers in radians.
[8 marks]
\({\text{B}}{{\text{C}}^2} + {({\text{BC}} + 1)^2} = 2\) M1A1
\(2{\text{B}}{{\text{C}}^2} + 2{\text{BC}} – 1 = 0\) A1
\({\text{BC}} = \frac{{ -2 + \sqrt {12} }}{4}\left( { = \frac{{\sqrt 3 – 1}}{2}} \right)\) M1A1
\(\sin \hat A = \frac{{{\text{BC}}}}{{\sqrt 2 }} = \frac{{\sqrt 3 – 1}}{{2\sqrt 2 }}\) A1
\( = \frac{{\sqrt 6 – \sqrt 2 }}{4}\) AG
[6 marks]
EITHER
\(h = {\text{ABsin}}\hat A\) M1
\( = ({\text{BC}} + 1)\sin \hat A\) A1
\( = \frac{{\sqrt 3 + 1}}{2} \times \frac{{\sqrt 6 – \sqrt 2 }}{4} = \frac{{\sqrt 2 }}{4}\) M1A1
OR
\(\tfrac{1}{2}AB.BC = \tfrac{1}{2}AC.h\) M1
\(\frac{{\sqrt 3 – 1}}{2} \cdot \frac{{\sqrt {3 + 1} }}{2} = \sqrt {2h} \) A1
\(\frac{2}{4} = \sqrt 2 h\) M1
\(h = \frac{1}{{2\sqrt 2 }}\) A1
[4 marks]
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