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Trigonometric Ratios and Identities · Paper 3 — Free Maths AA HL Practice Questions

1ChallengeLAQDefinitions of sine, cosine, and tangents9 marksPaper 3~14 min

Data

(ii) Hence, using the Squeeze Theorem, prove that

\lim_{x \to 0^+} \frac{\sin x}{x} = 1. \quad [4]$$ (iii) State why the result in (ii) extends to $\lim_{x \to 0} \dfrac{\sin x}{x} = 1$. $\quad [1]

You may use without proof that

\cos x \to 1

x \to 0

, and that

\sin^2 x + \cos^2 x = 1

A function

g

is defined by

g(x) = \dfrac{\sin x}{x}

for

x \neq 0

(a)

Using the unit circle below, where

0 < x < \dfrac{\pi}{2}

, the following areas can be established:

\text{Area of } \triangle OAP \leq \text{Area of sector } OAP \leq \text{Area of } \triangle OAT

where

O

is the origin,

A = (1,0)

P = (\cos x, \sin x)

lies on the unit circle, and

T

is the point where the tangent to the circle at

A

meets the line

OP

extended. (i) Show that the three areas satisfy $$\frac{1}{2}\sin x \leq \frac{x}{2} \leq \frac{1}{2}\tan x. \quad [3 marks]

(b)

Using the result from (a), evaluate

\lim_{x \to 0} \frac{1 - \cos x}{x^2},

justifying each step of your reasoning. $\quad [6 marks]

Solutions

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