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Example Question - trigonometry

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Find the Value of x in Degrees and Minutes

The image does not provide the full question, only a partial instruction which translates to "Find the value of x in degrees and minutes for the given." Without the full context or values, we cannot solve for x. To solve a question like this with given values, we would usually convert x from degrees to minutes by multiplying the degrees by 60, since there are 60 minutes in a degree. If x were given in minutes and seconds, we would convert the minutes to degrees by dividing by 60, and convert the seconds to minutes by dividing by 60 again, then add these to the minutes portion of the angle. However, the specific numeric value of x and the rest of the question is needed to provide a numerical answer.

Finding the Equation of a Line with a Given Point and Angle

The equation of a line in point-slope form is given by $y - y_1 = m(x - x_1)$, where $m$ is the slope of the line and $(x_1, y_1)$ is a point on the line. We are given the point $(1, 2)$ and that the line makes a $30^\circ$ angle with the x-axis. The slope $m$ of the line is the tangent of the angle it makes with the x-axis, thus $m = \tan(30^\circ) = \frac{\sqrt{3}}{3}$. Substituting the point and the slope into the point-slope form equation, we get: $y - 2 = \frac{\sqrt{3}}{3}(x - 1)$ Multiplying both sides by 3 to clear the fraction, we get: $3(y - 2) = \sqrt{3}(x - 1)$ Expanding, we have: $3y - 6 = \sqrt{3}x - \sqrt{3}$ Adding 6 and $\sqrt{3}$ to both sides gives us the final equation: $3y = \sqrt{3}x + 6 - \sqrt{3}$ Or we can express it in standard form $Ax + By = C$ by rearranging terms: $\sqrt{3}x - 3y = -6 + \sqrt{3}$

Intersection and Union of Sets, Trigonometric Functions, Complex Numbers, and Permutations

1. Let A be the set of those who teach physics, B the set of those who teach mathematics, and C the set of those who teach both. We can use the principle of inclusion and exclusion to find the number of teachers who teach either physics or mathematics. Let $ n(A) $ denote the number of physics teachers, $ n(B) $ the number of math teachers, and $ n(C) $ the number of teachers who teach both physics and mathematics. We are given $ n(A \cup B \cup C) = 12 $. We are also given that $ n(A \cup B) = n(A \cup C) $. To find $ n(A \cup B) $, we will calculate using inclusion-exclusion principle: $ n(A \cup B) = n(A) + n(B) - n(A \cap B) $ But since $ A \cap B = C $, we have: $ n(A \cup B) = n(A) + n(B) - n(C) $ Similarly, for $ n(A \cup C) $, assuming no one teaches only math, we have: $ n(A \cup C) = n(A) + n(C) - n(A \cap C) $ And since $ A \cap C = C $, we have: $ n(A \cup C) = n(A) $ Given that $ n(A \cup B) = n(A \cup C) $, we can set the equations equal to each other and solve: $ n(A) + n(B) - n(C) = n(A) $ $ n(B) - n(C) = 0 $ $ n(B) = n(C) $ Now substituting the values we have: $ 7 + n(B) - 4 = 12 $ $ n(B) - 4 = 5 $ $ n(B) = 9 $ $ n(A) = 12 - n(B) = 12 - 9 = 3 $ 2. To find $ B $ and $ R $ we use the following definitions: $ B = \{y | y = 2x + 7, x \in \mathbb{R} \text{ and } -5 \leq x \leq 5 \} $ The range of $ B $ is: When $ x = -5 $, $ y = 2(-5) + 7 = -3 $ When $ x = 5 $, $ y = 2(5) + 7 = 17 $ So, $ R(B) = [-3, 17] $ 3. If $ \sin(\theta) = -\frac{4}{5} $ and $ \theta $ is in the third quadrant, both $ \sin(\theta) $ and $ \cos(\theta) $ are negative in the third quadrant. To find $ \cos(\theta) $, we use the Pythagorean identity $ \sin^2(\theta) + \cos^2(\theta) = 1 $: $ \cos(\theta) = -\sqrt{1 - \sin^2(\theta)} = -\sqrt{1 - \left(-\frac{4}{5}\right)^2} = -\frac{3}{5} $ So, the real numbers $ x' $ and $ y' $ for $ \frac{\sin(\theta)}{\cos(\theta)} $ are: $ \frac{\sin(\theta)}{\cos(\theta)} = \frac{-\frac{4}{5}}{-\frac{3}{5}} = \frac{4}{3} $ Therefore, $ x' = 4 $ and $ y' = 3 $. 4. The word "INVOLUTE" has 3 vowels and 5 consonants. We can form different sets of 3 vowels and 2 consonants, and treat each set as a separate case for permutation. The number of permutations of 3 vowels (I, O, U) and 2 consonants out of 5 is given by: $ P(vowels) = \frac{3!}{(3-3)!} = 3! = 6 $ $ P(consonants) = \frac{5!}{(5-2)!} = \frac{5!}{3!} = 20 $ The total number of permutations is the product of the two: $ P(total) = P(vowels) \times P(consonants) = 6 \times 20 = 120 $ Each of these permutations can be arranged in 5! ways because the order of letters within the set matters. So, the total number of words is $ P(total) \times 5! = 120 \times 5! = 120 \times 120 = 14400 $.

Trigonometric Function Problem Involving Sine in Quadrants

Given that $\sin \theta = -\frac{3}{4}$ and $\theta$ lies in the third quadrant, we know that $\sin \theta < 0$ and $\cos \theta < 0$. To find $ \cos ^2 \theta $, let's first find $ \cos \theta $. We use the Pythagorean identity $ \sin^2 \theta + \cos^2 \theta = 1 $. $ \cos^2 \theta = 1 - \sin^2 \theta $ $ \cos^2 \theta = 1 - \left(-\frac{3}{4}\right)^2 = 1 - \frac{9}{16} = \frac{16}{16} - \frac{9}{16} = \frac{7}{16} $. The cosine function is negative in the third quadrant, so $\cos \theta = -\sqrt{\frac{7}{16}} = -\frac{\sqrt{7}}{4}$. Therefore, $ \cos^2 \theta = \left(-\frac{\sqrt{7}}{4}\right)^2 = \frac{7}{16} $.

Calculation of Trigonometric Expression and Centroid Coordinates

\text{For the trigonometric expression:} \sin(105^\circ) + \cos(105^\circ) = \sin(60^\circ+45^\circ) + \cos(60^\circ+45^\circ)\ \text{Using the sine and cosine addition formulas:} \sin(A+B) = \sin{A}\cos{B} + \cos{A}\sin{B} \cos(A+B) = \cos{A}\cos{B} - \sin{A}\sin{B} \text{Substitute } A = 60^\circ \text{ and } B = 45^\circ: \sin(60^\circ)\cos(45^\circ) + \cos(60^\circ)\sin(45^\circ) + \cos(60^\circ)\cos(45^\circ) - \sin(60^\circ)\sin(45^\circ)\ \text{Using known values of sine and cosine for } 60^\circ \text{ and } 45^\circ: \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2} + \frac{1}{2}\cdot\frac{\sqrt{2}}{2} + \frac{1}{2}\cdot\frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2} \text{Simplify the expression:} \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} + \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4} \text{The terms with } \sqrt{6} \text{ cancel out, so we are left with:} \frac{\sqrt{2}}{4} + \frac{\sqrt{2}}{4} = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2} \text{For the centroid coordinates:} \text{Let the coordinates of } C \text{ be } (x_C, y_C). \text{ The centroid coordinates (G) can be found by:} G_x = \frac{x_A + x_B + x_C}{3}, G_y = \frac{y_A + y_B + y_C}{3} \text{Substitute the given values with } G (1,1,1): 1 = \frac{3 + (-1) + x_C}{3}, 1 = \frac{-5 + (-7) + y_C}{3} \text{Solve for } x_C \text{ and } y_C: x_C = 3 - (3 + (-1)) = 3 - 2 = 1 y_C = 3 - (-5 -7) = 3 + 12 = 15 \text{So, the coordinates of point C are } (1, 15). \text{For the trigonometric identity:} \cot{(\theta)} + \tan{(\theta)} = 2\csc{(\theta)} \text{Using the identities } \cot{(\theta)} = \frac{1}{\tan{(\theta)}} \text{ and } \csc{(\theta)} = \frac{1}{\sin{(\theta)}}, \text{ we get:} \frac{1}{\tan{(\theta)}} + \tan{(\theta)} = \frac{2}{\sin{(\theta)}} \text{Multiply by } \tan{(\theta)}\sin{(\theta)} \text{ to clear the denominators:} \sin{(\theta)} + \tan^2{(\theta)}\sin{(\theta)} = 2\tan{(\theta)} \text{Use the identity } \sin^2{(\theta)} + \cos^2{(\theta)} = 1 \text{ to express } \tan{(\theta)} \text{ in terms of } \sin{(\theta)}: \tan{(\theta)} = \frac{\sin{(\theta)}}{\cos{(\theta)}}, \text{ so } \sin{(\theta)} + \frac{\sin^3{(\theta)}}{\cos^2{(\theta)}} = 2\frac{\sin{(\theta)}}{\cos{(\theta)}} \text{Multiply by } \cos^2{(\theta)} \text{ to clear the denominators:} \sin{(\theta)}\cos^2{(\theta)} + \sin^3{(\theta)} = 2\sin{(\theta)}\cos{(\theta)} \text{Rearrange and factor out } \sin{(\theta)}: \sin{(\theta)}(\cos^2{(\theta)} - 2\cos{(\theta)} + \sin^2{(\theta)}) = 0 \text{Since } \sin{(\theta)} = 0 \text{ leads to no solution that satisfies the original equation, focus on:} \cos^2{(\theta)} - 2\cos{(\theta)} + \sin^2{(\theta)} = 0 \text{Using } \sin^2{(\theta)} = 1 - \cos^2{(\theta)}: \cos^2{(\theta)} - 2\cos{(\theta)} + (1 - \cos^2{(\theta)}) = 0 -2\cos{(\theta)} + 1 = 0 \cos{(\theta)} = \frac{1}{2} \text{Hence, the general values of } \theta \text{ can be } 60^\circ + n360^\circ \text{ or } 300^\circ + n360^\circ, \text{ where } n \text{ is an integer.}

Inverse Tangent of a Rational Function

The given expression is: \[\tan^{-1}\left(\frac{1+x}{1-x}\right)\] Let's set $y = \tan^{-1}\left(\frac{1+x}{1-x}\right)$. To find the value of $\tan(y)$, we use the fact that $\tan(\tan^{-1}(x)) = x$. \[\tan(y) = \frac{1+x}{1-x}\] We can represent $y$ as the angle in a right triangle where the opposite side is $1+x$ and the adjacent side is $1-x$. To find the hypotenuse $h$, we use the Pythagorean theorem: \[h = \sqrt{(1+x)^2 + (1-x)^2}\] \[h = \sqrt{1+2x+x^2 + 1-2x+x^2}\] \[h = \sqrt{2x^2 + 2}\] \[h = \sqrt{2(x^2 + 1)}\] Now we express $\sin(y)$ and $\cos(y)$ in terms of the sides of the triangle: \[\sin(y) = \frac{opposite}{hypotenuse} = \frac{1+x}{\sqrt{2(x^2 + 1)}}\] \[\cos(y) = \frac{adjacent}{hypotenuse} = \frac{1-x}{\sqrt{2(x^2 + 1)}}\] We use the identity $\tan(y) = \frac{\sin(y)}{\cos(y)}$ to simplify: \[\tan(y) = \frac{\frac{1+x}{\sqrt{2(x^2 + 1)}}}{\frac{1-x}{\sqrt{2(x^2 + 1)}}}\] \[\tan(y) = \frac{1+x}{1-x}\] Which confirms our original expression. To find derivatives or perform integration, we would need additional context, as the problem does not specify an operation to perform with the given expression.

Right Triangle Problem

Para resolver este problema, necesitamos aplicar el teorema de Pitágoras, que establece que en un triángulo rectángulo la suma de los cuadrados de los catetos es igual al cuadrado de la hipotenusa. La fórmula es: \[a^2 + b^2 = c^2\] Donde $a$ y $b$ son los catetos, y $c$ es la hipotenusa. En este caso, tenemos un triángulo con catetos de longitud 6 cm y 7.5 cm. Aplicando el teorema de Pitágoras obtenemos: \[6^2 + 7.5^2 = c^2\] \[36 + 56.25 = c^2\] \[92.25 = c^2\] Tomando la raíz cuadrada de ambos lados: \[c = \sqrt{92.25}\] \[c = 9.6\, \text{cm}\] Por lo tanto, la longitud de la hipotenusa es de 9.6 cm. Para encontrar el valor del ángulo que no es de 90 grados y es adyacente al lado de 6 cm, podemos usar la función trigonométrica tangente, que es la relación entre el cateto opuesto y el cateto adyacente: \[\tan(\theta) = \frac{\text{cateto opuesto}}{\text{cateto adyacente}}\] En este caso, el cateto opuesto al ángulo es de 7.5 cm y el cateto adyacente es de 6 cm. Así que: \[\tan(\theta) = \frac{7.5}{6}\] \[\theta = \arctan\left(\frac{7.5}{6}\right)\] Usando una calculadora, encontramos que: \[\theta \approx 51.34^\circ\] Por lo tanto, el valor del ángulo $\theta$ es aproximadamente 51.34 grados.

Solving for the Sine Given the Tangent

Si tenemos que $\tan(\alpha) = 0.25$, recordemos que $\tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)}$. Ahora, podemos usar la identidad fundamental de la trigonometría que dice que $ \sin^2(\alpha) + \cos^2(\alpha) = 1 $. Despejando $\cos(\alpha)$, tenemos: $ \cos^2(\alpha) = 1 - \sin^2(\alpha) $. Pero como $\tan(\alpha) = 0.25 = \frac{\sin(\alpha)}{\cos(\alpha)} $, si despejamos $\sin(\alpha)$ en términos de $\cos(\alpha)$, obtenemos: $ \sin(\alpha) = 0.25 \cos(\alpha) $. Cuadrando ambos lados de la igualdad: $ \sin^2(\alpha) = 0.0625 \cos^2(\alpha) $. Reemplazando esto en la identidad fundamental, tenemos: $ 0.0625 \cos^2(\alpha) + \cos^2(\alpha) = 1 $. Sumando los términos semejantes: $ (0.0625 + 1) \cos^2(\alpha) = 1 $. $ 1.0625 \cos^2(\alpha) = 1 $. Despejando $\cos^2(\alpha)$, obtenemos: $ \cos^2(\alpha) = \frac{1}{1.0625} $. $ \cos^2(\alpha) = \frac{16}{17} $. Tomando la raíz cuadrada de ambos lados: $ \cos(\alpha) = \sqrt{\frac{16}{17}} $ o $ \cos(\alpha) = -\sqrt{\frac{16}{17}} $, pero como estamos buscando el seno, y la relación entre seno y coseno es positiva para el primer y tercer cuadrante, tomaremos la raíz positiva para encontrar el seno positivo correspondiente. $ \cos(\alpha) = \sqrt{\frac{16}{17}} $. Usando nuestra relación original entre el seno y el coseno: $ \sin(\alpha) = 0.25 \cdot \sqrt{\frac{16}{17}} $. $ \sin(\alpha) = \frac{0.25 \cdot 4}{\sqrt{17}} $. $ \sin(\alpha) = \frac{1}{\sqrt{17}} $. $ \sin(\alpha) = \sqrt{\frac{1}{17}} $. $ \sin(\alpha) ≈ \sqrt{0.05882352941} $. $ \sin(\alpha) ≈ 0.242535625 $.

Differentiation of a Composite Function with Trigonometric Factor

Используем правило производной произведения $(uv)'=u'v+uv'$ и производной степенной функции и тригонометрической функции. Дано: $f(x) = (4x^4 + 2)\cos(x)$. Найдем производную от $4x^4$: $(4x^4)' = 16x^3$. Производная от $\cos(x)$: $(\cos(x))' = -\sin(x)$. По правилу производной произведения: $f'(x) = (4x^4 + 2)' \cdot \cos(x) + (4x^4 + 2) \cdot (\cos(x))'$ $f'(x) = (16x^3) \cdot \cos(x) + (4x^4 + 2) \cdot (-\sin(x))$ $f'(x) = 16x^3\cos(x) - (4x^4 + 2)\sin(x)$

Approximating Values of Sine Using Right Triangle Ratios

Para la parte a), usaremos la relación trigonométrica del seno, que es el lado opuesto sobre la hipotenusa en un triángulo rectángulo. \[\sen(\alpha) = \frac{opuesto}{hipotenusa}\] \[\sen(\alpha) = \frac{2\ mm}{4\ mm}\] \[\sen(\alpha) = 0.5\] Para la parte b), de igual manera, aplicamos la relación del seno: \[\sen(\alpha) = \frac{opuesto}{hipotenusa}\] \[\sen(\alpha) = \frac{4\ cm}{\sqrt{10}\ cm}\] \[\sen(\alpha) = \frac{4}{\sqrt{10}}\] \[\sen(\alpha) = \frac{4}{\sqrt{10}} \cdot \frac{\sqrt{10}}{\sqrt{10}}\] \[\sen(\alpha) = \frac{4\sqrt{10}}{10}\] \[\sen(\alpha) = \frac{2\sqrt{10}}{5}\] Para obtener el valor aproximado a tres dígitos decimales, calculamos: \[\sen(\alpha) \approx \frac{2\sqrt{10}}{5} \approx 0.632\]

Trigonometric Ratio Calculation from Triangle Dimensions

Para el inciso a: \[ \sin(\alpha) = \frac{opuesto}{hipotenusa} = \frac{BC}{AB} = \frac{2 \text{ mm}}{4 \text{ mm}} = \frac{1}{2} \] Para el inciso b: \[ \sin(\alpha) = \frac{opuesto}{hipotenusa} = \frac{BC}{AB} = \frac{4 \text{ cm}}{\sqrt{10} \text{ cm}} = \frac{4}{\sqrt{10}} = \frac{4\sqrt{10}}{10} = \frac{2\sqrt{10}}{5} \]

Trigonometric Functions in Right Triangles

Para el inciso a, para calcular $x$ usamos la definición de seno en un triángulo rectángulo: $ \text{sen}(\theta) = \frac{\text{opuesto}}{\text{hipotenusa}} $. Entonces, tenemos que $ \text{sen}(\theta) = 0.28 = \frac{x}{24} $. $ x = 0.28 \cdot 24 $ $ x = 6.72 \text{ cm} $ Para el inciso b, para calcular la longitud del lado opuesto al ángulo $ \beta $, usaremos la definición de coseno: $ \text{cos}(\theta) = \frac{\text{adyacente}}{\text{hipotenusa}} $. Tenemos $ \text{cos}(\beta) = 0.324 = \frac{\text{adyacente}}{35} $. La longitud del lado adyacente es entonces $ \text{adyacente} = 0.324 \cdot 35 $. $ \text{adyacente} = 11.34 \text{ cm} $ Finalmente, para el inciso c, para calcular la longitud de $ \overline{AB} $, utilizamos la definición de tangente: $ \text{tg}(\theta) = \frac{\text{opuesto}}{\text{adyacente}} $. $ \text{tg}(\alpha) = 1.3 = \frac{\text{opuesto}}{10} $. La longitud del lado opuesto a $ \alpha $ es $ \text{opuesto} = 1.3 \cdot 10 $. $ \text{opuesto} = 13 \text{ cm} $

Trigonometry Problems Involving Right Triangles

Para el inciso a: $\text{Usamos la definición de seno en un triángulo rectángulo:} \frac{\text{Cateto opuesto}}{\text{Hipotenusa}}$ $sen(\theta) = \frac{x}{24}$ $0.28 = \frac{x}{24}$ $x = 0.28 \times 24$ $x = 6.72 \text{ cm}$ Para el inciso b: $\text{Usamos la definición de coseno en un triángulo rectángulo:} \frac{\text{Cateto adyacente}}{\text{Hipotenusa}}$ $cos(\beta) = \frac{35}{x}$ $0.324 = \frac{35}{x}$ $x = \frac{35}{0.324}$ $x \approx 108.025 \text{ cm}$ Para el inciso c: $\text{Usamos la definición de tangente en un triángulo rectángulo:} \frac{\text{Cateto opuesto}}{\text{Cateto adyacente}}$ $tg(\alpha) = \frac{x}{10}$ $1.3 = \frac{x}{10}$ $x = 1.3 \times 10$ $x = 13 \text{ cm}$

Trigonometric Functions in Right Triangles

Para la parte a), usando la definición de seno en un triángulo rectángulo: \[\text{sen}(\theta) = \frac{\text{opuesto}}{\text{hipotenusa}}\] Si $\text{sen}(\theta) = 0,28$, entonces la hipotenusa es $\frac{24}{0,28}$: \[x = \frac{24}{0,28}\] \[x = 85,71\ (2 d.p.)\] Para la parte b), usando la definición de coseno: \[\cos(\theta) = \frac{\text{adyacente}}{\text{hipotenusa}}\] Si $\cos(\beta) = 0,324$, entonces la longitud adyacente es $\frac{35}{0,324}$: \[x = \frac{35}{0,324}\] \[x = 108,02\ (2 d.p.)\] Para la parte c), usando la definición de tangente: \[\text{tg}(\theta) = \frac{\text{opuesto}}{\text{adyacente}}\] Si $\text{tg}(\alpha) = 1,3$, entonces la longitud adyacente es $\frac{10}{1,3}$: \[x = \frac{10}{1,3}\] \[x = 7,69\ (2 d.p.)\]

Solving for a Triangle Side Using Trigonometric Ratios

De acuerdo con la información proporcionada y utilizando la relación trigonométrica del seno, podemos encontrar el valor de $ x $. La fórmula para el seno de un ángulo en un triángulo rectángulo es: \[ \sin(\theta) = \frac{\text{opuesto}}{\text{hipotenusa}} \] Dado que $ \sin(\alpha) = 0.28 $ y la hipotenusa del triángulo es 8 unidades, podemos establecer la siguiente ecuación: \[ \sin(\alpha) = \frac{x}{8} \] Entonces: \[ 0.28 = \frac{x}{8} \] \[ x = 0.28 \cdot 8 \] \[ x = 2.24 \] Por lo tanto, el valor de $ x $ es 2.24 unidades.

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Example Question - trigonometry

Here are examples of questions we've helped users solve.

Find the Value of x in Degrees and Minutes

Finding the Equation of a Line with a Given Point and Angle

Intersection and Union of Sets, Trigonometric Functions, Complex Numbers, and Permutations

Trigonometric Function Problem Involving Sine in Quadrants

Calculation of Trigonometric Expression and Centroid Coordinates

Inverse Tangent of a Rational Function

Right Triangle Problem

Solving for the Sine Given the Tangent

Differentiation of a Composite Function with Trigonometric Factor

Approximating Values of Sine Using Right Triangle Ratios

Trigonometric Ratio Calculation from Triangle Dimensions

Trigonometric Functions in Right Triangles

Trigonometry Problems Involving Right Triangles

Trigonometric Functions in Right Triangles

Solving for a Triangle Side Using Trigonometric Ratios

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