Mid-Chapter Mixed Review

Determine whether the statement is true or false.

$sin x (csc x - cot x) = 1 - cos x$ $sin x (csc x - cot x) = 1 - cos x$ [7.1]
$sin 42 ° = \sqrt{\frac{1 + cos 84 °}{2}}$ $sin 42 ° = \sqrt{\frac{1 + cos 84 °}{2}}$ [7.2]
$sin \frac{π}{9} = cos \frac{7 π}{18}$ $sin \frac{π}{9} = cos \frac{7 π}{18}$ [7.2]
${cos}^{2} x \neq cos x^{2}$ ${cos}^{2} x \neq cos x^{2}$ [7.1]

For Exercises 5–14, choose one of expressions A–J to complete the identity. [7.1], [7.2]

Simplify.

$\sqrt{\frac{cot x}{sin x}}$ $\sqrt{\frac{cot x}{sin x}}$ [7.1]
$\frac{1}{{sin}^{2} x} - {(\frac{cos x}{sin x})}^{2}$ $\frac{1}{{sin}^{2} x} - {(\frac{cos x}{sin x})}^{2}$ [7.1]
$\frac{2 {cos}^{2} x - 5 cos x - 3}{cos x - 3}$ $\frac{2 {cos}^{2} x - 5 cos x - 3}{cos x - 3}$ [7.1]
$\frac{sin x}{tan (- x)}$ $\frac{sin x}{tan (- x)}$ [7.1]
${(cos x - sin x)}^{2}$ ${(cos x - sin x)}^{2}$ [7.2]
$1 - 2 {sin}^{2} \frac{x}{2}$ $1 - 2 {sin}^{2} \frac{x}{2}$ [7.2]
Rationalize the denominator: $\sqrt{\frac{sec x}{1 - cos x}} .$ $\sqrt{\frac{sec x}{1 - cos x}} .$ [7.1]
Write $cos 41 ° cos 29 ° + sin 41 ° sin 29 °$ $cos 41 ° cos 29 ° + sin 41 ° sin 29 °$ as a trigonometric function of a single angle and then evaluate.[7.1]
Evaluate $cos \frac{3 π}{8}$ $cos \frac{3 π}{8}$ exactly.[7.1]
Evaluate $sin 105 °$ $sin 105 °$ exactly.[7.1]
Assume that $sin α = \frac{5}{13}$ $sin α = \frac{5}{13}$ and $sin β = \frac{12}{13}$ $sin β = \frac{12}{13}$ and that $α$ $α$ and $β$ $β$ are between 0 and $π / 2$ $π / 2$ , and evaluate $tan (α - β)$ $tan (α - β)$ .[7.1]
Find the exact value of $sin 2 θ$ $sin 2 θ$ and the quadrant in which $2 θ$ $2 θ$ lies if $cos θ = - \frac{4}{5}$ $cos θ = - \frac{4}{5}$ , with $θ$ $θ$ in quadrant II.[7.2]

Prove the identity. [7.3]

${cos}^{2} \frac{x}{2} = \frac{tan x + sin x}{2 tan x}$ ${cos}^{2} \frac{x}{2} = \frac{tan x + sin x}{2 tan x}$
$\frac{1 - sin x}{cos x} = \frac{cos x}{1 + sin x}$ $\frac{1 - sin x}{cos x} = \frac{cos x}{1 + sin x}$
$\frac{{sin}^{3} x - {cos}^{3} x}{sin x - cos x} = \frac{2 + sin 2 x}{2}$ $\frac{{sin}^{3} x - {cos}^{3} x}{sin x - cos x} = \frac{2 + sin 2 x}{2}$
$sin 6 θ - sin 2 θ = tan 2 θ (cos 2 θ + cos 6 θ)$ $sin 6 θ - sin 2 θ = tan 2 θ (cos 2 θ + cos 6 θ)$

Collaborative Discussion and Writing

Explain why $tan (x + 450 °)$ $tan (x + 450 °)$ cannot be simplified using the tangent sum formula, but can be simplified using the sine and cosine sum formulas. [7.1]
Discuss and compare the graphs of $y = sin x$ $y = sin x$ , $y = sin 2 x$ $y = sin 2 x$ , and $y = sin (x / 2)$ $y = sin (x / 2)$ . [7.2]
What restrictions must be placed on the variable in each of the following identities? Why? [7.3]
1. a) $sin 2 x = \frac{2 tan x}{1 + {tan}^{2} x}$ $sin 2 x = \frac{2 tan x}{1 + {tan}^{2} x}$
2. b) $\frac{1 - cos x}{sin x} = \frac{sin x}{1 + cos x}$ $\frac{1 - cos x}{sin x} = \frac{sin x}{1 + cos x}$
Find all errors in the following:

$2 {sin}^{2} 2 x + cos 4 x$ $2 {sin}^{2} 2 x + cos 4 x$

$\begin{matrix} = & 2 {(2 sin x cos x)}^{2} + 2 cos 2 x \\ = & 8 {sin}^{2} x {cos}^{2} x + 2 ({cos}^{2} x + {sin}^{2} x) \\ = & 8 {sin}^{2} x {cos}^{2} x + 2. \end{matrix}$ $\begin{matrix} = & 2 {(2 sin x cos x)}^{2} + 2 cos 2 x \\ = & 8 {sin}^{2} x {cos}^{2} x + 2 ({cos}^{2} x + {sin}^{2} x) \\ = & 8 {sin}^{2} x {cos}^{2} x + 2. \end{matrix}$

[7.2]

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