Home / IB DP Maths / IB Math Analysis and Approach HL / MAA HL Study Notes / Definition of cos , sin and tan angles Study Notes

IB Mathematics AA Definition of cos , sin and tan angles Study Notes | New Syllabus

IB Mathematics AA Definition of cos , sin and tan angles Study Notes

IB Mathematics AA Definition of cos , sin and tan angles Study Notes

IB Mathematics AA Definition of cos , sin and tan angles Study Notes Offer a clear explanation of Definition of cos , sin and tan angles , 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 Definition of cos , sin and tan angles.

Definition of cos, sin, and tan

Trigonometric functions are fundamental in mathematics, defined geometrically and algebraically. This section explains the definitions of cosine, sine, and tangent using the unit circle and explores their relationships in different quadrants.

Key Concepts

Definition on the Unit Circle:

    • \(\cos\theta\): The x-coordinate of the point on the unit circle corresponding to the angle \( \theta \).
    • \(\sin\theta\): The y-coordinate of the point on the unit circle corresponding to the angle \( \theta \).
    • \(\tan\theta = \frac{\sin\theta}{\cos\theta}\).

Relationships in Different Quadrants:

    • \(\cos(-x) = \cos x\)
    • \(\tan(3\pi – x) = -\tan x\)
    • \(\sin(\pi + x) = -\sin x\)

Equation of a Straight Line:

    • The equation of a line through the origin is \(y = x \tan\theta\), where \( \theta \) is the angle between the line and the positive x-axis.

Exact Trigonometric Values:

    • \(\sin 0 = 0\), \(\cos 0 = 1\), \(\tan 0 = 0\).
    • \(\sin \frac{\pi}{6} = \frac{1}{2}, \, \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}, \, \tan \frac{\pi}{6} = \frac{\sqrt{3}}{3}\).
    • \(\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}, \, \cos \frac{\pi}{3} = \frac{1}{2}, \, \tan \frac{\pi}{3} = \sqrt{3}\).
    • \(\sin \frac{\pi}{2} = 1, \, \cos \frac{\pi}{2} = 0, \, \tan \frac{\pi}{2}\) is undefined.

Guidance, Clarifications, and Syllabus Links

  • Definitions and relationships should be supported with diagrams of the unit circle.
  • Connections to coordinate geometry, straight-line equations, and the symmetry of trigonometric functions across quadrants.
  • Key focus:
    • Use exact values for trigonometric ratios when solving problems.
    • Explore transformations of trigonometric functions on the unit circle.

Connections and Enrichment

Historical Context:

    • The Aryabhatiya of Aryabhata (ca. 510) was the first work to explicitly reference the sine as a function of an angle.

Theory of Knowledge (TOK):

    • Trigonometry was developed by successive civilizations and cultures. How have key events in the history of mathematics shaped its current form?
    • To what extent is mathematical knowledge embedded in particular traditions or bound to particular cultures?

Examples

Example 1: Quadrant Relationships

Find \(\cos(-\pi/4)\):

    • \(\cos(-\pi/4) = \cos(\pi/4) = \frac{\sqrt{2}}{2}\).

Example 2: Equation of a Line

Find the equation of a line through the origin making an angle of \(45^\circ\) with the x-axis:

    • \(y = x \tan 45^\circ = x \cdot 1 = x\).

IB Mathematics AA SL Definition of cos, sin, and tan Style Worked Out Questions

Question

(i)     Sketch the graphs of \(y = \sin x\) and \(y = \sin 2x\) , on the same set of axes, for \(0 \leqslant x \leqslant \frac{\pi }{2}\) .

(ii)     Find the x-coordinates of the points of intersection of the graphs in the domain \(0 \leqslant x \leqslant \frac{\pi }{2}\) .

(iii)     Find the area enclosed by the graphs.[9]

a.

Find the value of \(\int_0^1 {\sqrt {\frac{x}{{4 – x}}} }{{\text{d}}x} \) using the substitution \(x = 4{\sin ^2}\theta \) .[8]

b.

The increasing function f satisfies \(f(0) = 0\) and \(f(a) = b\) , where \(a > 0\) and \(b > 0\) .

(i)     By reference to a sketch, show that \(\int_0^a {f(x){\text{d}}x = ab – \int_0^b {{f^{ – 1}}(x){\text{d}}x} } \) .

(ii)     Hence find the value of \(\int_0^2 {\arcsin \left( {\frac{x}{4}} \right){\text{d}}x} \) .[8]

c.
▶️Answer/Explanation

Markscheme

(i)

    A2

Note: Award A1 for correct \(\sin x\) , A1 for correct \(\sin 2x\) .

 

Note: Award A1A0 for two correct shapes with \(\frac{\pi }{2}\) and/or 1 missing.

 

Note: Condone graph outside the domain.

(ii)     \(\sin 2x = \sin x\) , \(0 \leqslant x \leqslant \frac{\pi }{2}\)

\(2\sin x\cos x – \sin x = 0\)     M1

\(\sin x(2\cos x – 1) = 0\)

\(x = 0,\frac{\pi }{3}\)     A1A1     N1N1

 

(iii)     area \( = \int_0^{\frac{\pi }{3}} {(\sin 2x – \sin x){\text{d}}x} \)     M1

Note: Award M1 for an integral that contains limits, not necessarily correct, with \(\sin x\) and \(\sin 2x\) subtracted in either order.

\( = \left[ { – \frac{1}{2}\cos 2x + \cos x} \right]_0^{\frac{\pi }{3}}\)     A1

\( = \left( { – \frac{1}{2}\cos \frac{{2\pi }}{3} + \cos \frac{\pi }{3}} \right) – \left( { – \frac{1}{2}\cos 0 + \cos 0} \right)\)     (M1)

\( = \frac{3}{4} – \frac{1}{2}\)

\( = \frac{1}{4}\)     A1

[9 marks]

a.

\(\int_0^1 {\sqrt {\frac{x}{{4 – x}}} } {\text{d}}x = \int_0^{\frac{\pi }{6}} {\sqrt {\frac{{4{{\sin }^2}\theta }}{{4 – 4{{\sin }^2}\theta }}}  \times 8\sin \theta \cos \theta {\text{d}}\theta } \)     M1A1A1 

Note: Award M1 for substitution and reasonable attempt at finding expression for dx in terms of \({\text{d}}\theta \) , first A1 for correct limits, second A1 for correct substitution for dx .

\(\int_0^{\frac{\pi }{6}} {8{{\sin }^2}\theta {\text{d}}\theta } \)     A1

\(\int_0^{\frac{\pi }{6}} {4 – 4\cos 2\theta {\text{d}}\theta } \)     M1

\( = [4\theta – 2\sin 2\theta ]_0^{\frac{\pi }{6}}\)     A1

\( = \left( {\frac{{2\pi }}{3} – 2\sin \frac{\pi }{3}} \right) – 0\)     (M1)

\( = \frac{{2\pi }}{3} – \sqrt 3 \)     A1

[8 marks]

b.

(i)     

     M1

from the diagram above

the shaded area \( = \int_0^a {f(x){\text{d}}x = ab – \int_0^b {{f^{ – 1}}(y){\text{d}}y} } \)     R1

\({ = ab – \int_0^b {{f^{ – 1}}(x){\text{d}}x} }\)     AG

 

(ii)     \(f(x) = \arcsin \frac{x}{4} \Rightarrow {f^{ – 1}}(x) = 4\sin x\)     A1

\(\int_0^2 {\arcsin \left( {\frac{x}{4}} \right){\text{d}}x = \frac{\pi }{3} – \int_0^{\frac{\pi }{6}} {4\sin x{\text{d}}x} } \)     M1A1A1

Note: Award A1 for the limit \(\frac{\pi }{6}\) seen anywhere, A1 for all else correct.

 

\( = \frac{\pi }{3} – [ – 4\cos x]_0^{\frac{\pi }{6}}\)     A1

\( = \frac{\pi }{3} – 4 + 2\sqrt 3 \)     A1

Note: Award no marks for methods using integration by parts.

 

[8 marks]

c.

Question

Let \(f(x) = \frac{{\sin 3x}}{{\sin x}} – \frac{{\cos 3x}}{{\cos x}}\).

For what values of x does \(f(x)\) not exist?[2]

a.

Simplify the expression \(\frac{{\sin 3x}}{{\sin x}} – \frac{{\cos 3x}}{{\cos x}}\).[5]

b.
▶️Answer/Explanation

Markscheme

\(\cos x = 0{\text{, }}\sin x = 0\)     (M1)

\(x = \frac{{n\pi }}{2},n \in \mathbb{Z}\)     A1

a.

EITHER

\(\frac{{\sin 3x\cos x – \cos 3x\sin x}}{{\sin x\cos x}}\)     M1     A1

\( = \frac{{\sin (3x – x)}}{{\frac{1}{2}\sin 2x}}\)     A1     A1

\( = 2\)     A1

 

OR

\(\frac{{\sin 2x\cos x + \cos 2x\sin x}}{{\sin x}} – \frac{{\cos 2x\cos x – \sin 2x\sin x}}{{\cos x}}\)     M1

\( = \frac{{2\sin x{{\cos }^2}x + 2{{\cos }^2}x\sin x – \sin x}}{{\sin x}} – \frac{{2{{\cos }^3}x – \cos x – {{\sin }^2}x\cos x}}{{\cos x}}\)     A1     A1

\( = 4{\cos ^2}x – 1 – 2{\cos ^2}x + 1 + 2{\sin ^2}x\)     A1
\( = 2{\cos ^2}x + 2{\sin ^2}x\)

\( = 2\)     A1

 

[5 marks]

b.

More resources for IB Mathematics AA SL

Scroll to Top