解答｜代数と三角法｜プリンピキア

Solution 1解答 1

\begin{matrix} \csc θ \cos θ \tan θ & = & (\frac{1}{\sin θ}) \cos θ (\frac{\sin θ}{\cos θ}) \\ = & \frac{\cos θ}{\sin θ} (\frac{\sin θ}{\cos θ}) \\ = & \frac{\sin θ \cos θ}{\sin θ \cos θ} \\ = & 1 \end{matrix}

Solution 2解答 2

\begin{matrix} \frac{\cot θ}{\csc θ} & = & \frac{\frac{\cos θ}{\sin θ}}{\frac{1}{\sin θ}} \\ = & \frac{\cos θ}{\sin θ} \cdot \frac{\sin θ}{1} \\ = & \cos θ \end{matrix}

Solution 3解答 3

$\begin{matrix} \frac{\sin^{2} θ - 1}{\tan θ \sin θ - \tan θ} & = & \frac{(\sin θ + 1) (\sin θ - 1)}{\tan θ (\sin θ - 1)} \\ = & \frac{\sin θ + 1}{\tan θ} \end{matrix}$

Solution 4解答 4

This is a difference of squares formula: $25 - 9 \sin^{2} θ = (5 - 3 \sin θ) (5 + 3 \sin θ) .$

これは二乗の差の公式である: $25 - 9 \sin^{2} θ = (5 - 3 \sin θ) (5 + 3 \sin θ)$

Solution 5解答 5

\begin{matrix} \frac{\cos θ}{1 + \sin θ} (\frac{1 - \sin θ}{1 - \sin θ}) & = & \frac{\cos θ (1 - \sin θ)}{1 - \sin^{2} θ} \\ = & \frac{\cos θ (1 - \sin θ)}{\cos^{2} θ} \\ = & \frac{1 - \sin θ}{\cos θ} \end{matrix}

Solution 1解答 1

$\frac{\sqrt{2} + \sqrt{6}}{4}$

Solution 2解答 2

$\frac{\sqrt{2} - \sqrt{6}}{4}$

Solution 3解答 3

$\frac{1 - \sqrt{3}}{1 + \sqrt{3}}$

Solution 4解答 4

$\cos (\frac{5 π}{14})$

Solution 5解答 5

\begin{matrix} \tan (π - θ) & = & \frac{\tan (π) - \tan θ}{1 + \tan (π) \tan θ} \\ = & \frac{0 - \tan θ}{1 + 0 \cdot \tan θ} \\ = & - \tan θ \end{matrix}

Solution 1解答 1

$\cos (2 α) = \frac{7}{32}$

Solution 2解答 2

$\cos^{4} θ - \sin^{4} θ = (\cos^{2} θ + \sin^{2} θ) (\cos^{2} θ - \sin^{2} θ) = \cos (2 θ)$

Solution 3解答 3

$\cos (2 θ) \cos θ = (\cos^{2} θ - \sin^{2} θ) \cos θ = \cos^{3} θ - \cos θ \sin^{2} θ$

Solution 4解答 4

$\begin{matrix} 10 \cos^{4} x & = & 10 {(\cos^{2} x)}^{2} \\ = & 10 {[\frac{1 + \cos (2 x)}{2}]}^{2} & {Substitute reduction formula for cos}^{2} x . \\ = & \frac{10}{4} [1 + 2 \cos (2 x) + \cos^{2} (2 x)] \\ = & \frac{10}{4} + \frac{10}{2} \cos (2 x) + \frac{10}{4} (\frac{1 + \cos 2 (2 x)}{2}) & {Substitute reduction formula for cos}^{2} x . \\ = & \frac{10}{4} + \frac{10}{2} \cos (2 x) + \frac{10}{8} + \frac{10}{8} \cos (4 x) \\ = & \frac{30}{8} + 5 \cos (2 x) + \frac{10}{8} \cos (4 x) \\ = & \frac{15}{4} + 5 \cos (2 x) + \frac{5}{4} \cos (4 x) \end{matrix}$

式中の英語：Substitute reduction formula for cos：余弦の次数を下げる公式を使う

Solution 5解答 5

$- \frac{2}{\sqrt{5}}$

Solution 1解答 1

$\frac{1}{2} (\cos 6 θ + \cos 2 θ)$

Solution 2解答 2

$\frac{1}{2} (\sin 2 x + \sin 2 y)$

Solution 3解答 3

$\frac{- 2 - \sqrt{3}}{4}$

Solution 4解答 4

$2 \sin (2 θ) \cos (θ)$

Solution 5解答 5

$\begin{matrix} \tan θ \cot θ - \cos^{2} θ & = & (\frac{\sin θ}{\cos θ}) (\frac{\cos θ}{\sin θ}) - \cos^{2} θ \\ = & 1 - \cos^{2} θ \\ = & \sin^{2} θ \end{matrix}$

Solution 1解答 1

$x = \frac{7 π}{6}, \frac{11 π}{6}$

Solution 2解答 2

$\frac{π}{3} \pm π k$

Solution 3解答 3

$θ \approx 1.7722 \pm 2 π k$ and $θ \approx 4.5110 \pm 2 π k$

$θ \approx 1.7722 \pm 2 π k$ と $θ \approx 4.5110 \pm 2 π k$

Solution 4解答 4

$\cos θ = - 1, θ = π$

Solution 5解答 5

$\frac{π}{2}, \frac{2 π}{3}, \frac{4 π}{3}, \frac{3 π}{2}$

Solution 1解答 1

All three functions, $F$ , $G$ , and $H$ , are even.

三つの関数 $F$ 、 $G$ 、と $H$ 、はどれも偶関数である。

This is because $F (- x) = \sin (- x) \sin (- x) = (- \sin x) (- \sin x) = \sin^{2} x = F (x)$ , $G (- x) = \cos (- x) \cos (- x) = \cos x \cos x = \cos^{2} x = G (x)$ and $H (- x) = \tan (- x) \tan (- x) = (- \tan x) (- \tan x) = \tan^{2} x = H (x) .$

これは $F (- x) = \sin (- x) \sin (- x) = (- \sin x) (- \sin x) = \sin^{2} x = F (x)$ 、 $G (- x) = \cos (- x) \cos (- x) = \cos x \cos x = \cos^{2} x = G (x)$ 、 $H (- x) = \tan (- x) \tan (- x) = (- \tan x) (- \tan x) = \tan^{2} x = H (x)$ だからである。

Solution 3解答 3

When $\cos t = 0,$ then $\sec t = \frac{1}{0},$ which is undefined.

$\cos t = 0$ のとき $\sec t = \frac{1}{0}$ となり、定義されない。

Solution 5解答 5

$\sin x$

Solution 7解答 7

$\sec x$

Solution 9解答 9

$\csc t$

Solution 11解答 11

$−1$

Solution 13解答 13

$\sec^{2} x$

Solution 15解答 15

$\sin^{2} x + 1$

Solution 17解答 17

$\frac{1}{\sin x}$

Solution 19解答 19

$\frac{1}{\cot x}$

Solution 21解答 21

$\tan x$

Solution 23解答 23

$- 4 \sec x \tan x$

Solution 25解答 25

$\pm \sqrt{\frac{1}{\cot^{2} x} + 1}$

Solution 27解答 27

$\frac{\pm \sqrt{1 - \sin^{2} x}}{\sin x}$

Solution 29解答 29

Answers will vary. Sample proof:

答えはさまざまでよい。証明の例:

$\begin{matrix} \cos x - \cos^{3} x & = & \cos x (1 - \cos^{2} x) \\ = & \cos x \sin^{2} x \end{matrix}$

Solution 31解答 31

Answers will vary. Sample proof:
$\frac{1 + \sin^{2} x}{\cos^{2} x} = \frac{1}{\cos^{2} x} + \frac{\sin^{2} x}{\cos^{2} x} = \sec^{2} x + \tan^{2} x = \tan^{2} x + 1 + \tan^{2} x = 1 + 2 \tan^{2} x$

答えはさまざまでよい。証明の例: $\frac{1 + \sin^{2} x}{\cos^{2} x} = \frac{1}{\cos^{2} x} + \frac{\sin^{2} x}{\cos^{2} x} = \sec^{2} x + \tan^{2} x = \tan^{2} x + 1 + \tan^{2} x = 1 + 2 \tan^{2} x$

Solution 33解答 33

Answers will vary. Sample proof:
$\cos^{2} x - \tan^{2} x = 1 - \sin^{2} x - (\sec^{2} x - 1) = 1 - \sin^{2} x - \sec^{2} x + 1 = 2 - \sin^{2} x - \sec^{2} x$

答えはさまざまでよい。証明の例: $\cos^{2} x - \tan^{2} x = 1 - \sin^{2} x - (\sec^{2} x - 1) = 1 - \sin^{2} x - \sec^{2} x + 1 = 2 - \sin^{2} x - \sec^{2} x$

Solution 35解答 35

False

正しくない

Solution 37解答 37

False

正しくない

Solution 39解答 39

Proved with negative and Pythagorean identities

負の等式とピタゴラスの等式で証明した。

Solution 41解答 41

True $3 \sin^{2} θ + 4 \cos^{2} θ = 3 \sin^{2} θ + 3 \cos^{2} θ + \cos^{2} θ = 3 (\sin^{2} θ + \cos^{2} θ) + \cos^{2} θ = 3 + \cos^{2} θ$

正しい $3 \sin^{2} θ + 4 \cos^{2} θ = 3 \sin^{2} θ + 3 \cos^{2} θ + \cos^{2} θ = 3 (\sin^{2} θ + \cos^{2} θ) + \cos^{2} θ = 3 + \cos^{2} θ$

Solution 1解答 1

The cofunction identities apply to complementary angles. Viewing the two acute angles of a right triangle, if one of those angles measures $x,$ the second angle measures $\frac{π}{2} - x .$ Then $\sin x = \cos (\frac{π}{2} - x) .$ The same holds for the other cofunction identities. The key is that the angles are complementary.

余関数の等式は余角に当てはまる。直角三角形の二つの鋭角を見て、一方が $x$ なら、もう一方は $\frac{π}{2} - x$ である。すると $\sin x = \cos (\frac{π}{2} - x)$ となる。ほかの余関数の等式でも同じである。要は、その角が互いに余角であることである。

Solution 3解答 3

$\sin (- x) = - \sin x,$ so $\sin x$ is odd. $\cos (- x) = \cos (0 - x) = \cos x,$ so $\cos x$ is even.

$\sin (- x) = - \sin x$ なので $\sin x$ は奇関数である。 $\cos (- x) = \cos (0 - x) = \cos x$ なので $\cos x$ は偶関数である。

Solution 5解答 5

$\frac{\sqrt{2} + \sqrt{6}}{4}$

Solution 7解答 7

$\frac{\sqrt{6} - \sqrt{2}}{4}$

Solution 9解答 9

$- 2 - \sqrt{3}$

Solution 11解答 11

$- \frac{\sqrt{2}}{2} \sin x - \frac{\sqrt{2}}{2} \cos x$

Solution 13解答 13

$- \frac{1}{2} \cos x - \frac{\sqrt{3}}{2} \sin x$

Solution 15解答 15

$\csc θ$

Solution 17解答 17

$\cot x$

Solution 19解答 19

$\tan (\frac{x}{10})$

Solution 21解答 21

$\begin{matrix} \sin (a - b) & = & (\frac{4}{5}) (\frac{1}{3}) - (\frac{3}{5}) (\frac{2 \sqrt{2}}{3}) & = & \frac{4 - 6 \sqrt{2}}{15} \\ \cos (a + b) & = & (\frac{3}{5}) (\frac{1}{3}) - (\frac{4}{5}) (\frac{2 \sqrt{2}}{3}) & = & \frac{3 - 8 \sqrt{2}}{15} \end{matrix}$

Solution 23解答 23

$\frac{\sqrt{2} - \sqrt{6}}{4}$

Solution 25解答 25

$\sin x$

Solution 27解答 27

$\cot (\frac{π}{6} - x)$

Graph of y=cot(pi/6 - x) from -2pi to pi - in comparison to the usual y=cot(x) graph, this one is reflected across the x-axis and shifted by pi/6.

Solution 29解答 29

$\cot (\frac{π}{4} + x)$

Graph of y=cot(pi/4 + x) - in comparison to the usual y=cot(x) graph, this one is shifted by pi/4.

Solution 31解答 31

$\frac{\sin x}{\sqrt{2}} + \frac{\cos x}{\sqrt{2}}$

Graph of y = sin(x) / rad2 + cos(x) / rad2 - it looks like the sin curve shifted by pi/4.

Solution 33解答 33

They are the same.

同じである。

Solution 35解答 35

They are the different, try $g (x) = \sin (9 x) - \cos (3 x) \sin (6 x) .$

違う。 $g (x) = \sin (9 x) - \cos (3 x) \sin (6 x)$ で試せ。

Solution 37解答 37

They are the same.

同じである。

Solution 39解答 39

They are the different, try $g (θ) = \frac{2 \tan θ}{1 - \tan^{2} θ} .$

違う。 $g (θ) = \frac{2 \tan θ}{1 - \tan^{2} θ}$ で試せ。

Solution 41解答 41

They are different, try $g (x) = \frac{\tan x - \tan (2 x)}{1 + \tan x \tan (2 x)} .$

違う。 $g (x) = \frac{\tan x - \tan (2 x)}{1 + \tan x \tan (2 x)}$ で試せ。

Solution 43解答 43

$- \frac{\sqrt{3} - 1}{2 \sqrt{2}}, or - 0.2588$

式中の英語：or：または

Solution 45解答 45

$\frac{1 + \sqrt{3}}{2 \sqrt{2}},$ or 0.9659

$\frac{1 + \sqrt{3}}{2 \sqrt{2}}$ すなわち0.9659

Solution 47解答 47

$\begin{matrix} \tan (x + \frac{π}{4}) & = \\ \frac{\tan x + \tan (\frac{π}{4})}{1 - \tan x \tan (\frac{π}{4})} & = \\ \frac{\tan x + 1}{1 - \tan x (1)} & = & \frac{\tan x + 1}{1 - \tan x} \end{matrix}$

Solution 49解答 49

$\begin{matrix} \frac{\cos (a + b)}{\cos a \cos b} & = \\ \frac{\cos a \cos b}{\cos a \cos b} - \frac{\sin a \sin b}{\cos a \cos b} & = & 1 - \tan a \tan b \end{matrix}$

Solution 51解答 51

$\begin{matrix} \frac{\cos (x + h) - \cos x}{h} & = \\ \frac{\cos x \cosh - \sin x \sinh - \cos x}{h} & = \\ \frac{\cos x (\cosh - 1) - \sin x \sinh}{h} & = & \cos x \frac{\cos h - 1}{h} - \sin x \frac{\sin h}{h} \end{matrix}$

Solution 53解答 53

True

正しい

Solution 55解答 55

True. Note that $\sin (α + β) = \sin (π - γ)$ and expand the right hand side.

正しい。 $\sin (α + β) = \sin (π - γ)$ に注意して右辺を展開せよ。

Solution 1解答 1

Use the Pythagorean identities and isolate the squared term.

ピタゴラスの等式を使って、2乗の項を取り出す。

Solution 3解答 3

$\frac{1 - \cos x}{\sin x}, \frac{\sin x}{1 + \cos x},$ multiplying the top and bottom by $\sqrt{1 - \cos x}$ and $\sqrt{1 + \cos x},$ respectively.

$\frac{1 - \cos x}{\sin x}, \frac{\sin x}{1 + \cos x}$ 。分子と分母にそれぞれ $\sqrt{1 - \cos x}$ と $\sqrt{1 + \cos x}$ を掛ける。

Solution 5解答 5

a) $\frac{3 \sqrt{7}}{32}$ b) $\frac{31}{32}$ c) $\frac{3 \sqrt{7}}{31}$

Solution 7解答 7

a) $\frac{\sqrt{3}}{2}$ b) $- \frac{1}{2}$ c) $- \sqrt{3}$

Solution 9解答 9

$\cos θ = - \frac{2 \sqrt{5}}{5}, \sin θ = \frac{\sqrt{5}}{5}, \tan θ = - \frac{1}{2}, \csc θ = \sqrt{5}, \sec θ = - \frac{\sqrt{5}}{2}, \cot θ = - 2$

Solution 11解答 11

$\sin (\frac{π}{2})$

Solution 13解答 13

$\frac{\sqrt{2 - \sqrt{2}}}{2}$

Solution 15解答 15

$\frac{\sqrt{2 - \sqrt{3}}}{2}$

Solution 17解答 17

$2 + \sqrt{3}$

Solution 19解答 19

$- 1 - \sqrt{2}$

Solution 21解答 21

a) $\frac{3 \sqrt{13}}{13}$ b) $- \frac{2 \sqrt{13}}{13}$ c) $- \frac{3}{2}$

Solution 23解答 23

a) $\frac{\sqrt{10}}{4}$ b) $\frac{\sqrt{6}}{4}$ c) $\frac{\sqrt{15}}{3}$

Solution 25解答 25

$\frac{120}{169}, - \frac{119}{169}, - \frac{120}{119}$

Solution 27解答 27

$\frac{2 \sqrt{13}}{13}, \frac{3 \sqrt{13}}{13}, \frac{2}{3}$

Solution 29解答 29

$\cos (74°)$

Solution 31解答 31

$\cos (18 x)$

Solution 33解答 33

$3 \sin (10 x)$

Solution 35解答 35

$- 2 \sin (- x) \cos (- x) = - 2 (- \sin (x) \cos (x)) = \sin (2 x)$

Solution 37解答 37

$\begin{matrix} \frac{\sin (2 θ)}{1 + \cos (2 θ)} \tan^{2} θ & = & \frac{2 \sin (θ) \cos (θ)}{1 + \cos^{2} θ - \sin^{2} θ} \tan^{2} θ = \\ \frac{2 \sin (θ) \cos (θ)}{2 \cos^{2} θ} \tan^{2} θ & = & \frac{\sin (θ)}{\cos θ} \tan^{2} θ = \\ \cot (θ) \tan^{2} θ & = & \tan^{3} θ \end{matrix}$

Solution 39解答 39

$\frac{1 + \cos (12 x)}{2}$

Solution 41解答 41

$\frac{3 + \cos (12 x) - 4 \cos (6 x)}{8}$

Solution 43解答 43

$\frac{2 + \cos (2 x) - 2 \cos (4 x) - \cos (6 x)}{32}$

Solution 45解答 45

$\frac{3 + \cos (4 x) - 4 \cos (2 x)}{3 + \cos (4 x) + 4 \cos (2 x)}$

Solution 47解答 47

$\frac{1 - \cos (4 x)}{8}$

Solution 49解答 49

$\frac{3 + \cos (4 x) - 4 \cos (2 x)}{4 (\cos (2 x) + 1)}$

Solution 51解答 51

$\frac{(1 + \cos (4 x)) \sin x}{2}$

Solution 53解答 53

$4 \sin x \cos x (\cos^{2} x - \sin^{2} x)$

Solution 55解答 55

$\begin{array}{l} \frac{2 \tan x}{1 + \tan^{2} x} = \frac{\frac{2 \sin x}{\cos x}}{1 + \frac{\sin^{2} x}{\cos^{2} x}} = \frac{\frac{2 \sin x}{\cos x}}{\frac{\cos^{2} x + \sin^{2} x}{\cos^{2} x}} = \\ \frac{2 \sin x}{\cos x} . \frac{\cos^{2} x}{1} = 2 \sin x \cos x = \sin (2 x) \end{array}$

Solution 57解答 57

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

Solution 59解答 59

$\begin{matrix} \sin (x + 2 x) & = & \sin x \cos (2 x) + \sin (2 x) \cos x \\ = & \sin x (\cos^{2} x - \sin^{2} x) + 2 \sin x \cos x \cos x \\ = & \sin x \cos^{2} x - \sin^{3} x + 2 \sin x \cos^{2} x \\ = & 3 \sin x \cos^{2} x - \sin^{3} x \end{matrix}$

Solution 61解答 61

$\begin{matrix} \frac{1 + \cos (2 t)}{\sin (2 t) - \cos t} & = & \frac{1 + 2 \cos^{2} t - 1}{2 \sin t \cos t - \cos t} \\ = & \frac{2 \cos^{2} t}{\cos t (2 \sin t - 1)} \\ = & \frac{2 \cos t}{2 \sin t - 1} \end{matrix}$

Solution 63解答 63

$\begin{matrix} (\cos^{2} (4 x) - \sin^{2} (4 x) - \sin (8 x)) (\cos^{2} (4 x) - \sin^{2} (4 x) + \sin (8 x)) & = \\ = & (\cos (8 x) - \sin (8 x)) (\cos (8 x) + \sin (8 x)) \\ = & \cos^{2} (8 x) - \sin^{2} (8 x) \\ = & \cos (16 x) \end{matrix}$

Solution 1解答 1

Substitute $α$ into cosine and $β$ into sine and evaluate.

余弦に $α$ 、正弦に $β$ を代入して値を求める。

Solution 3解答 3

Answers will vary. There are some equations that involve a sum of two trig expressions where when converted to a product are easier to solve. For example: $\frac{\sin (3 x) + \sin x}{\cos x} = 1.$ When converting the numerator to a product the equation becomes: $\frac{2 \sin (2 x) \cos x}{\cos x} = 1$

答えはさまざまでよい。二つの三角関数の式の和を含む方程式の中には、積に直すと解きやすくなるものがある。たとえば $\frac{\sin (3 x) + \sin x}{\cos x} = 1.$ 。分子を積に直すと、この方程式は次のようになる: $\frac{2 \sin (2 x) \cos x}{\cos x} = 1$ 。

Solution 5解答 5

$8 (\cos (5 x) - \cos (27 x))$

Solution 7解答 7

$\sin (2 x) + \sin (8 x)$

Solution 9解答 9

$\frac{1}{2} (\cos (6 x) - \cos (4 x))$

Solution 11解答 11

$2 \cos (5 t) \cos t$

Solution 13解答 13

$2 \cos (7 x)$

Solution 15解答 15

$2 \cos (6 x) \cos (3 x)$

Solution 17解答 17

$\frac{1}{4} (1 + \sqrt{3})$

Solution 19解答 19

$\frac{1}{4} (\sqrt{3} - 2)$

Solution 21解答 21

$\frac{1}{4} (\sqrt{3} - 1)$

Solution 23解答 23

$\cos (80°) - \cos (120°)$

Solution 25解答 25

$\frac{1}{2} (\sin (221°) + \sin (205°))$

Solution 27解答 27

$\sqrt{2} \cos (31°)$

Solution 29解答 29

$2 \cos (66.5°) \sin (34.5°)$

Solution 31解答 31

$2 \sin (−1.5°) \cos (0.5°)$

Solution 33解答 33

$\begin{array}{l} 2 \sin (7 x) - 2 \sin x = 2 \sin (4 x + 3 x) - 2 \sin (4 x - 3 x) = \\ 2 (\sin (4 x) \cos (3 x) + \sin (3 x) \cos (4 x)) - 2 (\sin (4 x) \cos (3 x) - \sin (3 x) \cos (4 x)) = \\ 2 \sin (4 x) \cos (3 x) + 2 \sin (3 x) \cos (4 x)) - 2 \sin (4 x) \cos (3 x) + 2 \sin (3 x) \cos (4 x)) = \\ 4 \sin (3 x) \cos (4 x) \end{array}$

Solution 35解答 35

$\begin{matrix} \sin x + \sin (3 x) & = & 2 \sin (\frac{4 x}{2}) \cos (\frac{- 2 x}{2}) = \\ 2 \sin (2 x) \cos x & = & 2 (2 \sin x \cos x) \cos x = \\ 4 \sin x \cos^{2} x \end{matrix}$

Solution 37解答 37

$\begin{array}{l} 2 \tan x \cos (3 x) = \frac{2 \sin x \cos (3 x)}{\cos x} = \frac{2 (.5 (\sin (4 x) - \sin (2 x)))}{\cos x} = \\ \frac{1}{\cos x} (\sin (4 x) - \sin (2 x)) = \sec x (\sin (4 x) - \sin (2 x)) \end{array}$

Solution 39解答 39

$2 \cos (35°) \cos (23°), 1.5081$

Solution 41解答 41

$- 2 \sin (33°) \sin (11°), - 0.2078$

Solution 43解答 43

$\frac{1}{2} (\cos (99°) - \cos (71°)), −0.2410$

Solution 45解答 45

It is an identity.

これは等式である。

Solution 47解答 47

It is not an identity, but $2 \cos^{3} x$ is.

これは等式でないが、 $2 \cos^{3} x$ は等式である。

Solution 49解答 49

$\tan (3 t)$

Solution 51解答 51

$2 \cos (2 x)$

Solution 53解答 53

$- \sin (14 x)$

Solution 55解答 55

Start with $\cos x + \cos y .$ Make a substitution and let $x = α + β$ and let $y = α - β,$ so $\cos x + \cos y$ becomes $\begin{array}{l} \cos (α + β) + \cos (α - β) = \cos α \cos β - \sin α \sin β + \cos α \cos β + \sin α \sin β = \\ 2 \cos α \cos β \end{array}$

$\cos x + \cos y$ から始める。置き換えをして $x = α + β$ 、 $y = α - β$ とおくと、 $\cos x + \cos y$ は $\begin{array}{l} \cos (α + β) + \cos (α - β) = \cos α \cos β - \sin α \sin β + \cos α \cos β + \sin α \sin β = \\ 2 \cos α \cos β \end{array}$ になる。

Since $x = α + β$ and $y = α - β,$ we can solve for $α$ and $β$ in terms of x and y and substitute in for $2 \cos α \cos β$ and get $2 \cos (\frac{x + y}{2}) \cos (\frac{x - y}{2}) .$

$x = α + β$ かつ $y = α - β$ なので、 $α$ と $β$ を x と y で解いて $2 \cos α \cos β$ に代入でき、 $2 \cos (\frac{x + y}{2}) \cos (\frac{x - y}{2})$ が得られる。

Solution 57解答 57

$\frac{\cos (3 x) + \cos x}{\cos (3 x) - \cos x} = \frac{2 \cos (2 x) \cos x}{- 2 \sin (2 x) \sin x} = - \cot (2 x) \cot x$

Solution 59解答 59

$\begin{matrix} \frac{\cos (2 y) - \cos (4 y)}{\sin (2 y) + \sin (4 y)} & = & \frac{- 2 \sin (3 y) \sin (- y)}{2 \sin (3 y) \cos y} = \\ \frac{2 \sin (3 y) \sin (y)}{2 \sin (3 y) \cos y} & = & \tan y \end{matrix}$

Solution 61解答 61

$\begin{array}{l} \cos x - \cos (3 x) = - 2 \sin (2 x) \sin (- x) = \\ 2 (2 \sin x \cos x) \sin x = 4 \sin^{2} x \cos x \end{array}$

Solution 63解答 63

$\tan (\frac{π}{4} - t) = \frac{\tan (\frac{π}{4}) - \tan t}{1 + \tan (\frac{π}{4}) \tan (t)} = \frac{1 - \tan t}{1 + \tan t}$

Solution 1解答 1

There will not always be solutions to trigonometric function equations. For a basic example, $\cos (x) = −5.$

三角関数の方程式に、つねに解があるわけではない。基本の例として $\cos (x) = −5.$ がある。

Solution 3解答 3

If the sine or cosine function has a coefficient of one, isolate the term on one side of the equals sign. If the number it is set equal to has an absolute value less than or equal to one, the equation has solutions, otherwise it does not. If the sine or cosine does not have a coefficient equal to one, still isolate the term but then divide both sides of the equation by the leading coefficient. Then, if the number it is set equal to has an absolute value greater than one, the equation has no solution.

正弦や余弦の関数の係数が1なら、その項を等号の片側に取り出す。それが等しいとおかれた数の絶対値が1以下なら、この方程式には解があり、そうでなければ解は無い。正弦や余弦の係数が1でないなら、やはりその項を取り出し、それから方程式の両辺を最高次係数で割る。そして、それが等しいとおかれた数の絶対値が1より大きいなら、この方程式に解は無い。

Solution 5解答 5

$\frac{π}{3}, \frac{2 π}{3}$

Solution 7解答 7

$\frac{3 π}{4}, \frac{5 π}{4}$

Solution 9解答 9

$\frac{π}{4}, \frac{5 π}{4}$

Solution 11解答 11

$\frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$

Solution 13解答 13

$\frac{π}{4}, \frac{7 π}{4}$

Solution 15解答 15

$\frac{7 π}{6}, \frac{11 π}{6}$

Solution 17解答 17

$\frac{π}{18}, \frac{5 π}{18}, \frac{13 π}{18}, \frac{17 π}{18}, \frac{25 π}{18}, \frac{29 π}{18}$

Solution 19解答 19

$\frac{3 π}{12}, \frac{5 π}{12}, \frac{11 π}{12}, \frac{13 π}{12}, \frac{19 π}{12}, \frac{21 π}{12}$

Solution 21解答 21

$\frac{1}{6}, \frac{5}{6}, \frac{13}{6}, \frac{17}{6}, \frac{25}{6}, \frac{29}{6}, \frac{37}{6}$

Solution 23解答 23

$0, \frac{π}{3}, π, \frac{5 π}{3}$

Solution 25解答 25

$\frac{π}{3}, π, \frac{5 π}{3}$

Solution 27解答 27

$\frac{π}{3}, \frac{3 π}{2}, \frac{5 π}{3}$

Solution 29解答 29

$0, π$

Solution 31解答 31

$π - \sin^{- 1} (- \frac{1}{4}), \frac{7 π}{6}, \frac{11 π}{6}, 2 π + \sin^{- 1} (- \frac{1}{4})$

Solution 33解答 33

$\frac{1}{3} (\sin^{- 1} (\frac{9}{10}))$ , $\frac{π}{3} - \frac{1}{3} (\sin^{- 1} (\frac{9}{10}))$ , $\frac{2 π}{3} + \frac{1}{3} (\sin^{- 1} (\frac{9}{10}))$ , $π - \frac{1}{3} (\sin^{- 1} (\frac{9}{10}))$ , $\frac{4 π}{3} + \frac{1}{3} (\sin^{- 1} (\frac{9}{10}))$ , $\frac{5 π}{3} - \frac{1}{3} (\sin^{- 1} (\frac{9}{10}))$

Solution 35解答 35

$0$

Solution 37解答 37

$\frac{π}{6}, \frac{5 π}{6}, \frac{7 π}{6}, \frac{11 π}{6}$

Solution 39解答 39

$\frac{3 π}{2}, \frac{π}{6}, \frac{5 π}{6}$

Solution 41解答 41

$0, \frac{π}{3}, π, \frac{4 π}{3}$

Solution 43解答 43

There are no solutions.

解は無い。

Solution 45解答 45

$\cos^{- 1} (\frac{1}{3} (1 - \sqrt{7}))$ , $2 π - \cos^{- 1} (\frac{1}{3} (1 - \sqrt{7}))$

Solution 47解答 47

$\tan^{- 1} (\frac{1}{2} (\sqrt{29} - 5))$ , $π + \tan^{- 1} (\frac{1}{2} (- \sqrt{29} - 5))$ , $π + \tan^{- 1} (\frac{1}{2} (\sqrt{29} - 5))$ , $2 π + \tan^{- 1} (\frac{1}{2} (- \sqrt{29} - 5))$

Solution 49解答 49

There are no solutions.

解は無い。

Solution 51解答 51

There are no solutions.

解は無い。

Solution 53解答 53

$0, \frac{2 π}{3}, \frac{4 π}{3}$

Solution 55解答 55

$\frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$

Solution 57解答 57

$\sin^{- 1} (\frac{3}{5}), \frac{π}{2}, π - \sin^{- 1} (\frac{3}{5}), \frac{3 π}{2}$

Solution 59解答 59

$\cos^{- 1} (- \frac{1}{4}), 2 π - \cos^{- 1} (- \frac{1}{4})$

Solution 61解答 61

$\frac{π}{3}$ , $\cos^{- 1} (- \frac{3}{4})$ , $2 π - \cos^{- 1} (- \frac{3}{4})$ , $\frac{5 π}{3}$

Solution 63解答 63

$\cos^{- 1} (\frac{3}{4})$ , $\cos^{- 1} (- \frac{2}{3})$ , $2 π - \cos^{- 1} (- \frac{2}{3})$ , $2 π - \cos^{- 1} (\frac{3}{4})$

Solution 65解答 65

$0, \frac{π}{2}, π, \frac{3 π}{2}$

Solution 67解答 67

$\frac{π}{3}$ , $\cos^{−1} (- \frac{1}{4})$ , $2 π - \cos^{−1} (- \frac{1}{4})$ , $\frac{5 π}{3}$

Solution 69解答 69

There are no solutions.

解は無い。

Solution 71解答 71

$π + \tan^{−1} (−2)$ , $π + \tan^{−1} (- \frac{3}{2})$ , $2 π + \tan^{−1} (−2)$ , $2 π + \tan^{−1} (- \frac{3}{2})$

Solution 73解答 73

$2 π k + 0.2734, 2 π k + 2.8682$

Solution 75解答 75

$π k - 0.3277$

Solution 77解答 77

$0.6694, 1.8287, 3.8110, 4.9703$

Solution 79解答 79

$1.0472, 3.1416, 5.2360$

Solution 81解答 81

$0.5326, 1.7648, 3.6742, 4.9064$

Solution 83解答 83

$\sin^{- 1} (\frac{1}{4}), π - \sin^{- 1} (\frac{1}{4}), \frac{3 π}{2}$

Solution 85解答 85

$\frac{π}{2}, \frac{3 π}{2}$

Solution 87解答 87

There are no solutions.

解は無い。

Solution 89解答 89

$0, \frac{π}{2}, π, \frac{3 π}{2}$

Solution 91解答 91

There are no solutions.

解は無い。

Solution 93解答 93

${7.2}^{\circ}$

Solution 95解答 95

${5.7}^{\circ}$

Solution 97解答 97

${82.4}^{\circ}$

Solution 99解答 99

${31.0}^{\circ}$

Solution 101解答 101

${88.7}^{\circ}$

Solution 103解答 103

${59.0}^{\circ}$

Solution 105解答 105

${36.9}^{\circ}$

Solution 1解答 1

$\sin^{- 1} (\frac{\sqrt{3}}{3})$ , $π - \sin^{- 1} (\frac{\sqrt{3}}{3})$ , $π + \sin^{- 1} (\frac{\sqrt{3}}{3})$ , $2 π - \sin^{- 1} (\frac{\sqrt{3}}{3})$

Solution 3解答 3

$\frac{7 π}{6}, \frac{11 π}{6}$

Solution 5解答 5

$\sin^{- 1} (\frac{1}{4}), π - \sin^{- 1} (\frac{1}{4})$

Solution 7解答 7

$1$

Solution 9解答 9

Yes

はい

Solution 11解答 11

$- 2 - \sqrt{3}$

Solution 13解答 13

$\frac{\sqrt{2}}{2}$

Solution 15解答 15

$\begin{matrix} \cos (4 x) - \cos (3 x) \cos x & = & \cos (2 x + 2 x) - \cos (x + 2 x) \cos x \\ = & \cos (2 x) \cos (2 x) - \sin (2 x) \sin (2 x) - \cos x \cos (2 x) \cos x + \sin x \sin (2 x) \cos x \\ = & {(\cos^{2} x - \sin^{2} x)}^{2} - 4 \cos^{2} x \sin^{2} x - \cos^{2} x (\cos^{2} x - \sin^{2} x) + \sin x (2) \sin x \cos x \cos x \\ = & {(\cos^{2} x - \sin^{2} x)}^{2} - 4 \cos^{2} x \sin^{2} x - \cos^{2} x (\cos^{2} x - \sin^{2} x) + 2 \sin^{2} x \cos^{2} x \\ = & \cos^{4} x - 2 \cos^{2} x \sin^{2} x + \sin^{4} x - 4 \cos^{2} x \sin^{2} x - \cos^{4} x + \cos^{2} x \sin^{2} x + 2 \sin^{2} x \cos^{2} x \\ = & \sin^{4} x - 4 \cos^{2} x \sin^{2} x + \cos^{2} x \sin^{2} x \\ = & \sin^{2} x (\sin^{2} x + \cos^{2} x) - 4 \cos^{2} x \sin^{2} x \\ = & \sin^{2} x - 4 \cos^{2} x \sin^{2} x \end{matrix}$

Solution 17解答 17

$\tan (\frac{5}{8} x)$

Solution 19解答 19

$\frac{\sqrt{3}}{3}$

Solution 21解答 21

$- \frac{24}{25}, - \frac{7}{25}, \frac{24}{7}$

Solution 23解答 23

$\sqrt{2 (2 + \sqrt{2})}$

Solution 25解答 25

$\frac{\sqrt{2}}{10}, \frac{7 \sqrt{2}}{10}, \frac{1}{7}, \frac{3}{5}, \frac{4}{5}, \frac{3}{4}$

Solution 27解答 27

$\begin{matrix} \cot x \cos (2 x) & = & \cot x (1 - 2 \sin^{2} x) \\ = & \cot x - \frac{\cos x}{\sin x} (2) \sin^{2} x \\ = & - 2 \sin x \cos x + \cot x \\ = & - \sin (2 x) + \cot x \end{matrix}$

Solution 29解答 29

$\frac{10 \sin x - 5 \sin (3 x) + \sin (5 x)}{8 (\cos (2 x) + 1)}$

Solution 31解答 31

$\frac{\sqrt{3}}{2}$

Solution 33解答 33

$- \frac{\sqrt{2}}{2}$

Solution 35解答 35

$\frac{1}{2} (\sin (6 x) + \sin (12 x))$

Solution 37解答 37

$2 \sin (\frac{13}{2} x) \cos (\frac{9}{2} x)$

Solution 39解答 39

$\frac{3 π}{4}, \frac{7 π}{4}$

Solution 41解答 41

$0, \frac{π}{6}, \frac{5 π}{6}, π$

Solution 43解答 43

$\frac{3 π}{2}$

Solution 45解答 45

No solution

解なし

Solution 47解答 47

$0.2527, 2.8889, 4.7124$

Solution 49解答 49

$1.3694, 1.9106, 4.3726, 4.9137$

Solution 1解答 1

1

Solution 3解答 3

$\sec (θ)$

Solution 5解答 5

$\frac{\sqrt{2} - \sqrt{6}}{4}$

Solution 7解答 7

$- 2 - \sqrt{3}$

Solution 9解答 9

$- \frac{1}{2} \cos θ + \frac{\sqrt{3}}{2} \sin θ$

Solution 11解答 11

$\frac{1 - \cos (64^{\circ})}{2}$

Solution 13解答 13

$0, π$

Solution 15解答 15

$\frac{π}{2}, \frac{3 π}{2}$

Solution 17解答 17

$2 cos (3 x) cos (5 x)$

Solution 19解答 19

$4 \sin (2 θ) \cos (6 θ)$

Solution 21解答 21

$x = {cos}^{–1} (\frac{1}{5})$

Solution 23解答 23

$\frac{π}{3}$

Solution 25解答 25

$\frac{3}{5}, - \frac{4}{5}, - \frac{3}{4}$

Solution 27解答 27

$\begin{matrix} {tan}^{3} x - tan x {sec}^{2} x & = & tan x ({tan}^{2} x - {sec}^{2} x) \\ = & tan x ({tan}^{2} x - (1 + {tan}^{2} x)) \\ = & tan x ({tan}^{2} x - 1 - {tan}^{2} x) \\ = & - tan x = tan (- x) = tan (- x) \end{matrix}$

Solution 29解答 29

$\begin{matrix} \frac{sin (2 x)}{sin x} - \frac{cos (2 x)}{cos x} & = & \frac{2 sin x cos x}{sin x} - \frac{2 {cos}^{2} x - 1}{cos x} \\ = & 2 cos x - 2 cos x + \frac{1}{cos x} \\ = & \frac{1}{cos x} = sec x = sec x \end{matrix}$

Solution 31解答 31

Amplitude: $\frac{1}{4}$ , period: $\frac{1}{60}$ , frequency: 60 Hz

振幅: $\frac{1}{4}$ 、周期: $\frac{1}{60}$ 、振動数: 60 Hz

Solution 33解答 33

Amplitude: 8, fast period: $\frac{1}{500}$ , fast frequency: 500 Hz, slow period: $\frac{1}{10}$ , slow frequency: 10 Hz

振幅: 8、速いほうの周期: $\frac{1}{500}$ 、速いほうの振動数: 500 Hz、遅いほうの周期: $\frac{1}{10}$ 、遅いほうの振動数: 10 Hz

Solution 35解答 35

$D (t) = 20 (0.9086)^{t} cos (4 π t)$ , 31 second

$D (t) = 20 (0.9086)^{t} cos (4 π t)$ 、31秒

第9章三角関数の恒等式と方程式Trigonometric Identities and Equations

9.1 Section Exercises9.1 節末問題

9.2 Section Exercises9.2 節末問題

9.3 Section Exercises9.3 節末問題

9.4 Section Exercises9.4 節末問題

9.5 Section Exercises9.5 節末問題

Review Exercises復習問題

Practice Test実力テスト