Example Question - sine
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Solving for the Sine Given the Tangent
<p>Si tenemos que \(\tan(\alpha) = 0.25\), recordemos que \(\tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)}\).</p> <p>Ahora, podemos usar la identidad fundamental de la trigonometría que dice que \( \sin^2(\alpha) + \cos^2(\alpha) = 1 \). Despejando \(\cos(\alpha)\), tenemos:</p> <p>\( \cos^2(\alpha) = 1 - \sin^2(\alpha) \).</p> <p>Pero como \(\tan(\alpha) = 0.25 = \frac{\sin(\alpha)}{\cos(\alpha)} \), si despejamos \(\sin(\alpha)\) en términos de \(\cos(\alpha)\), obtenemos:</p> <p>\( \sin(\alpha) = 0.25 \cos(\alpha) \).</p> <p>Cuadrando ambos lados de la igualdad:</p> <p>\( \sin^2(\alpha) = 0.0625 \cos^2(\alpha) \).</p> <p>Reemplazando esto en la identidad fundamental, tenemos:</p> <p>\( 0.0625 \cos^2(\alpha) + \cos^2(\alpha) = 1 \).</p> <p>Sumando los términos semejantes:</p> <p>\( (0.0625 + 1) \cos^2(\alpha) = 1 \).</p> <p>\( 1.0625 \cos^2(\alpha) = 1 \).</p> <p>Despejando \(\cos^2(\alpha)\), obtenemos:</p> <p>\( \cos^2(\alpha) = \frac{1}{1.0625} \).</p> <p>\( \cos^2(\alpha) = \frac{16}{17} \).</p> <p>Tomando la raíz cuadrada de ambos lados:</p> <p>\( \cos(\alpha) = \sqrt{\frac{16}{17}} \) 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.</p> <p>\( \cos(\alpha) = \sqrt{\frac{16}{17}} \).</p> <p>Usando nuestra relación original entre el seno y el coseno:</p> <p>\( \sin(\alpha) = 0.25 \cdot \sqrt{\frac{16}{17}} \).</p> <p>\( \sin(\alpha) = \frac{0.25 \cdot 4}{\sqrt{17}} \).</p> <p>\( \sin(\alpha) = \frac{1}{\sqrt{17}} \).</p> <p>\( \sin(\alpha) = \sqrt{\frac{1}{17}} \).</p> <p>\( \sin(\alpha) ≈ \sqrt{0.05882352941} \).</p> <p>\( \sin(\alpha) ≈ 0.242535625 \).</p>
Trigonometric Expression Simplification
<p>Для решения данной задачи, используйте тригонометрические формулы.</p> <p>Упростим выражение \(\frac{1 - \cos(2\alpha)}{\sin(2\alpha)}\), используя тригонометрические тождества:</p> <p>\(1 - \cos(2\alpha) = 1 - (1 - 2\sin^2(\alpha)) = 2\sin^2(\alpha)\)</p> <p>\(\sin(2\alpha) = 2\sin(\alpha)\cos(\alpha)\)</p> <p>Теперь подставим упрощённые выражения:</p> <p>\(\frac{1 - \cos(2\alpha)}{\sin(2\alpha)} = \frac{2\sin^2(\alpha)}{2\sin(\alpha)\cos(\alpha)}\)</p> <p>Сократите выражение на \(2\sin(\alpha)\):</p> <p>\(\frac{2\sin^2(\alpha)}{2\sin(\alpha)\cos(\alpha)} = \frac{\sin(\alpha)}{\cos(\alpha)}\)</p> <p>Используя определение тангенса, получаем:</p> <p>\(\frac{\sin(\alpha)}{\cos(\alpha)} = \tan(\alpha)\)</p> <p>Итак, \(\frac{1 - \cos(2\alpha)}{\sin(2\alpha)} = \tan(\alpha)\)</p>
Approximating Values of Sine Using Right Triangle Ratios
<p>Para la parte a), usaremos la relación trigonométrica del seno, que es el lado opuesto sobre la hipotenusa en un triángulo rectángulo.</p> <p>\[\sen(\alpha) = \frac{opuesto}{hipotenusa}\]</p> <p>\[\sen(\alpha) = \frac{2\ mm}{4\ mm}\]</p> <p>\[\sen(\alpha) = 0.5\]</p> <p>Para la parte b), de igual manera, aplicamos la relación del seno:</p> <p>\[\sen(\alpha) = \frac{opuesto}{hipotenusa}\]</p> <p>\[\sen(\alpha) = \frac{4\ cm}{\sqrt{10}\ cm}\]</p> <p>\[\sen(\alpha) = \frac{4}{\sqrt{10}}\]</p> <p>\[\sen(\alpha) = \frac{4}{\sqrt{10}} \cdot \frac{\sqrt{10}}{\sqrt{10}}\]</p> <p>\[\sen(\alpha) = \frac{4\sqrt{10}}{10}\]</p> <p>\[\sen(\alpha) = \frac{2\sqrt{10}}{5}\]</p> <p>Para obtener el valor aproximado a tres dígitos decimales, calculamos:</p> <p>\[\sen(\alpha) \approx \frac{2\sqrt{10}}{5} \approx 0.632\]</p>
Trigonometric Functions in Right Triangles
<p>Para el inciso a, para calcular \(x\) usamos la definición de seno en un triángulo rectángulo: \( \text{sen}(\theta) = \frac{\text{opuesto}}{\text{hipotenusa}} \).</p> <p>Entonces, tenemos que \( \text{sen}(\theta) = 0.28 = \frac{x}{24} \).</p> <p>\( x = 0.28 \cdot 24 \)</p> <p>\( x = 6.72 \text{ cm} \)</p> <p>Para el inciso b, para calcular la longitud del lado opuesto al ángulo \( \beta \), usaremos la definición de coseno: \( \text{cos}(\theta) = \frac{\text{adyacente}}{\text{hipotenusa}} \).</p> <p>Tenemos \( \text{cos}(\beta) = 0.324 = \frac{\text{adyacente}}{35} \).</p> <p>La longitud del lado adyacente es entonces \( \text{adyacente} = 0.324 \cdot 35 \).</p> <p>\( \text{adyacente} = 11.34 \text{ cm} \)</p> <p>Finalmente, para el inciso c, para calcular la longitud de \( \overline{AB} \), utilizamos la definición de tangente: \( \text{tg}(\theta) = \frac{\text{opuesto}}{\text{adyacente}} \).</p> <p>\( \text{tg}(\alpha) = 1.3 = \frac{\text{opuesto}}{10} \).</p> <p>La longitud del lado opuesto a \( \alpha \) es \( \text{opuesto} = 1.3 \cdot 10 \).</p> <p>\( \text{opuesto} = 13 \text{ cm} \)</p>
Trigonometry Problems Involving Right Triangles
Para el inciso a: <p>\(\text{Usamos la definición de seno en un triángulo rectángulo:} \frac{\text{Cateto opuesto}}{\text{Hipotenusa}}\)</p> <p>\(sen(\theta) = \frac{x}{24}\)</p> <p>\(0.28 = \frac{x}{24}\)</p> <p>\(x = 0.28 \times 24\)</p> <p>\(x = 6.72 \text{ cm}\)</p> Para el inciso b: <p>\(\text{Usamos la definición de coseno en un triángulo rectángulo:} \frac{\text{Cateto adyacente}}{\text{Hipotenusa}}\)</p> <p>\(cos(\beta) = \frac{35}{x}\)</p> <p>\(0.324 = \frac{35}{x}\)</p> <p>\(x = \frac{35}{0.324}\)</p> <p>\(x \approx 108.025 \text{ cm}\)</p> Para el inciso c: <p>\(\text{Usamos la definición de tangente en un triángulo rectángulo:} \frac{\text{Cateto opuesto}}{\text{Cateto adyacente}}\)</p> <p>\(tg(\alpha) = \frac{x}{10}\)</p> <p>\(1.3 = \frac{x}{10}\)</p> <p>\(x = 1.3 \times 10\)</p> <p>\(x = 13 \text{ cm}\)</p>
Trigonometric Functions in Right Triangles
<p>Para la parte a), usando la definición de seno en un triángulo rectángulo:</p> <p>\[\text{sen}(\theta) = \frac{\text{opuesto}}{\text{hipotenusa}}\]</p> <p>Si \(\text{sen}(\theta) = 0,28\), entonces la hipotenusa es \(\frac{24}{0,28}\):</p> <p>\[x = \frac{24}{0,28}\]</p> <p>\[x = 85,71\ (2 d.p.)\]</p> <p>Para la parte b), usando la definición de coseno:</p> <p>\[\cos(\theta) = \frac{\text{adyacente}}{\text{hipotenusa}}\]</p> <p>Si \(\cos(\beta) = 0,324\), entonces la longitud adyacente es \(\frac{35}{0,324}\):</p> <p>\[x = \frac{35}{0,324}\]</p> <p>\[x = 108,02\ (2 d.p.)\]</p> <p>Para la parte c), usando la definición de tangente:</p> <p>\[\text{tg}(\theta) = \frac{\text{opuesto}}{\text{adyacente}}\]</p> <p>Si \(\text{tg}(\alpha) = 1,3\), entonces la longitud adyacente es \(\frac{10}{1,3}\):</p> <p>\[x = \frac{10}{1,3}\]</p> <p>\[x = 7,69\ (2 d.p.)\]</p>
Solving for a Triangle Side Using Trigonometric Ratios
<p>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:</p> <p>\[ \sin(\theta) = \frac{\text{opuesto}}{\text{hipotenusa}} \]</p> <p>Dado que \( \sin(\alpha) = 0.28 \) y la hipotenusa del triángulo es 8 unidades, podemos establecer la siguiente ecuación:</p> <p>\[ \sin(\alpha) = \frac{x}{8} \]</p> <p>Entonces:</p> <p>\[ 0.28 = \frac{x}{8} \]</p> <p>\[ x = 0.28 \cdot 8 \]</p> <p>\[ x = 2.24 \]</p> <p>Por lo tanto, el valor de \( x \) es 2.24 unidades.</p>
Solving for the Sine of an Angle in Similar Triangles
<p>Given that \(\sin(F) = \frac{308}{317}\) and triangles FGH and JKL are similar with angle F corresponding to angle J, and angles G and K are right angles, we have:</p> <p>\(\sin(J) = \sin(F) = \frac{308}{317}\)</p> <p>Therefore, the value of \(\sin(J)\) is \(\frac{308}{317}\).</p> <p>The correct option is C) \(\frac{308}{317}\).</p>
Solving for the Sine of Angle R in a Right Triangle Given the Tangent of Angle S
Given that \(\tan(S) = \frac{1}{\sqrt{3}}\) and \(\angle T = 90\) degrees, so \(\angle S + \angle R = 90\) degrees. \(\tan(S) = \frac{1}{\sqrt{3}} = \frac{\text{opposite}}{\text{adjacent}} = \frac{\sin(S)}{\cos(S)}\) \(\sin(S) = \cos(R)\) since \(\sin(\theta) = \cos(90 - \theta)\) Therefore, \(\sin(S) = \frac{1}{\sqrt{3}}\) Now, since \(\sin(S) = \frac{1}{\sqrt{3}}\), we have \(\sin(R) = \sin(90 - S)\) \(\sin(R) = \cos(S) = \sqrt{1 - \sin^2(S)}\) Calculating \(\cos(S)\) given \(\sin(S)\): \(\cos(S) = \sqrt{1 - \left(\frac{1}{\sqrt{3}}\right)^2}\) \(\cos(S) = \sqrt{1 - \frac{1}{3}} = \sqrt{\frac{2}{3}}\) \(\sin(R) = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{2} \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}}\) \(\sin(R) = \frac{\sqrt{6}}{3}\) Since this option is not provided in the multiple-choice answers, it must be simplified further: \(\sin(R) = \frac{\sqrt{6}}{3} = \frac{2 \cdot \sqrt{6}}{2 \cdot 3}\) \(\sin(R) = \frac{2\sqrt{6}}{6}\) \(\sin(R) = \frac{\sqrt{6}}{3}\) Hence, the correct answer is \(\sin(R) = \frac{\sqrt{6}}{3}\), which is not listed in the provided options, there might be an error in the provided options or question setup.
Trigonometric Functions Calculation
写真に写っている問題を解くために、次のように進めます。 与えられた方程式は次の通りです。 \[ \sin \alpha = \frac{1}{3}, \quad \cos \beta = -\frac{2}{5} \] 求めるべきは次の式の値です。 \[ \sin(\alpha - \beta), \quad \cos(\alpha + \beta) \] まず初めに、与えられた条件から、余弦定理や正弦定理を使わずに三角関数の値を求めるために、以下のピタゴラスの定理に基づく基本的な関係を用います。 \[ \sin^2 \theta + \cos^2 \theta = 1 \] この関係を使うことで、与えられた \(\sin \alpha\) と \(\cos \beta\) の値から、残りの三角関数の値を導き出すことが可能です。 \[ \sin \alpha = \frac{1}{3} \] \[ \sin^2 \alpha= \left(\frac{1}{3}\right)^2 = \frac{1}{9} \] \[ \cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{1}{9} = \frac{8}{9} \] \[ \cos \alpha = \pm \sqrt{\frac{8}{9}} = \pm \frac{2\sqrt{2}}{3} \] 同様に、 \[ \cos \beta = -\frac{2}{5} \] \[ \cos^2 \beta = \left(-\frac{2}{5}\right)^2 = \frac{4}{25} \] \[ \sin^2 \beta = 1 - \cos^2 \beta = 1 - \frac{4}{25} = \frac{21}{25} \] \[ \sin \beta = \pm \sqrt{\frac{21}{25}} = \pm \frac{\sqrt{21}}{5} \] 三角関数の正負については、角 \(\alpha\) と \(\beta\) がどの象限にあるかによって決まりますが、この問題にはそのような情報が与えられていません。ですので、正負の両方のケースを検討する必要があります。しかし、通常、この手の問題では、最も一般的な第1象限（すなわち、全ての三角関数の値が正）を仮定します。 三角関数の加法定理に基づき、次の式を使用します。 \[ \sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \] \[ \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \] 得られた三角関数の値を以上の式に適用します（ここでは正の値を使用します）。 まずは \(\sin(\alpha - \beta)\) から求めます。 \[ \sin(\alpha - \beta) = \left(\frac{1}{3}\right)\left(-\frac{2}{5}\right) - \left(\frac{2\sqrt{2}}{3}\right)\left(\frac{\sqrt{21}}{5}\right) \] \[ \sin(\alpha - \beta) = -\frac{2}{15} - \frac{2\sqrt{42}}{15} = -\frac{2(1 + \sqrt{42})}{15} \] 次に \(\cos(\alpha + \beta)\) を求めます。 \[ \cos(\alpha + \beta) = \left(\frac{2\sqrt{2}}{3}\right)\left(-\frac{2}{5}\right) - \left(\frac{1}{3}\right)\left(\frac{\sqrt{21}}{5}\right) \] \[ \cos(\alpha + \beta) = -\frac{4\sqrt{2}}{15} - \frac{\sqrt{21}}{15} = -\frac{4\sqrt{2} + \sqrt{21}}{15} \] 以上の計算を踏まえて、最終的な結果は以下の通りになります。 \[ \sin(\alpha - \beta) = -\frac{2(1 + \sqrt{42})}{15} \] \[ \cos(\alpha + \beta) = -\frac{4\sqrt{2} + \sqrt{21}}{15} \]
Trigonometric Ratios in the Second Quadrant
The question gives us the cosine of an angle (θ) and the range for the angle, which is the second quadrant (\(\frac{\pi}{2} < \theta < \pi\)), and asks us to find the values of the remaining trigonometric ratios. Given: \[ \cos \theta = -\frac{\sqrt{3}}{2} \] Since the cosine of θ is negative and the angle is in the second quadrant, sine will be positive (as sine is positive in the second quadrant). Let's use the Pythagorean identity to find sine: \[ \sin^2 \theta + \cos^2 \theta = 1 \] Substitute the given value of cos θ: \[ \sin^2 \theta + \left(-\frac{\sqrt{3}}{2}\right)^2 = 1 \] \[ \sin^2 \theta + \frac{3}{4} = 1 \] Solving for \(\sin^2 \theta\): \[ \sin^2 \theta = 1 - \frac{3}{4} = \frac{1}{4} \] Taking the square root to find sin θ yields two possible values, +1/2 and -1/2. Since we are in the second quadrant and sine is positive: \[ \sin \theta = \frac{1}{2} \] The tangent of θ is the ratio of sine to cosine: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \] \[ \tan \theta = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} \] \[ \tan \theta = -\frac{1}{\sqrt{3}} \] \[ \tan \theta = -\frac{\sqrt{3}}{3} \] For the reciprocals: \[ \csc \theta = \frac{1}{\sin \theta} = 2 \] \[ \sec \theta = \frac{1}{\cos \theta} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3} \] \[ \cot \theta = \frac{1}{\tan \theta} = -\sqrt{3} \] So, the six trigonometric ratios for the given angle θ in the second quadrant are: \[ \sin \theta = \frac{1}{2}, \cos \theta = -\frac{\sqrt{3}}{2}, \tan \theta = -\frac{\sqrt{3}}{3} \] \[ \csc \theta = 2, \sec \theta = -\frac{2\sqrt{3}}{3}, \cot \theta = -\sqrt{3} \]
Solving Complex Number Division
Certainly! To find \( \frac{z_1}{z_2} \) given that \( z_1 = \cos \theta + j \sin \theta \) and \( z_2 = \cos \theta - j \sin \theta \), we can directly divide the complex numbers as follows: \[ \frac{z_1}{z_2} = \frac{\cos \theta + j \sin \theta}{\cos \theta - j \sin \theta} \] To divide these two complex numbers, we can multiply the numerator and the denominator by the conjugate of the denominator to remove the imaginary parts from the denominator. The conjugate of \( z_2 \) is \( \cos \theta + j \sin \theta \), which is actually the same as \( z_1 \). So, we multiply both top and bottom by this conjugate: \[ \frac{z_1}{z_2} = \frac{(\cos \theta + j \sin \theta)(\cos \theta + j \sin \theta)}{(\cos \theta - j \sin \theta)(\cos \theta + j \sin \theta)} \] Multiplying out the numerator, we get: Numerator: \( (\cos^2 \theta + 2j \cos \theta \sin \theta - \sin^2 \theta) \) (using \( j^2 = -1 \)) Multiplying out the denominator, we get: Denominator: \( (\cos^2 \theta - (\sin^2 \theta)) \) Simplify the denominator using \( \cos^2 \theta - \sin^2 \theta = \cos(2\theta) \): Denominator: \( \cos(2\theta) \) However, the denominator simplifies even further because \( \cos^2 \theta - \sin^2 \theta = 1 \) (from the Pythagorean identity \( \cos^2 \theta + \sin^2 \theta = 1 \)). Therefore, the denominator actually is: Denominator: \( 1 \) Returning to the numerator, we use the identity \( \cos^2 \theta + \sin^2 \theta = 1 \) again: Numerator: \( 1 + 2j \cos \theta \sin \theta \) The numerator simplifies using the double angle identity for sine: \( 2\sin \theta \cos \theta = \sin(2\theta) \): Numerator: \( 1 + j \sin(2\theta) \) Putting it all together, we have: \[ \frac{z_1}{z_2} = 1 + j \sin(2\theta) \] This is the final simplified form of \( \frac{z_1}{z_2} \).
Trigonometry Question: Finding Cosine Value Given Sine
The image shows a trigonometry question which reads: "Question 26 You are told that sin θ = \(\frac{7}{\sqrt{389}}\). a) If θ is in the first quadrant, then the exact value of cos θ is __________. Note: In this question we require you input your answer without decimals and without entering the words sin, cos or tan. For example, if your answer is \(\frac{\sqrt{17}}{17}\), then enter sqrt(5)/sqrt(17)" To solve for cos θ when given sin θ, we can use the Pythagorean identity, which states that sin² θ + cos² θ = 1. Since we're given sin θ = \(\frac{7}{\sqrt{389}}\), we can square this to find sin² θ: (sin θ)² = \(\left(\frac{7}{\sqrt{389}}\right)^2 = \frac{49}{389}\) Next we'll use the identity to solve for cos² θ: cos² θ = 1 - sin² θ = 1 - \(\frac{49}{389}\) To find cos² θ, we need to subtract \(\frac{49}{389}\) from 1. Since 1 can be written as \(\frac{389}{389}\), we get: cos² θ = \(\frac{389}{389} - \frac{49}{389} = \frac{389 - 49}{389} = \frac{340}{389}\) So, cos² θ = \(\frac{340}{389}\). We are now looking for the positive square root since θ is in the first quadrant where cosine values are positive: cos θ = \(\sqrt{\frac{340}{389}}\) The square root of a fraction is the square root of the numerator over the square root of the denominator: cos θ = \(\frac{\sqrt{340}}{\sqrt{389}}\) In the most simplified fractional form, this is the exact value for cos θ. However, note that \(\sqrt{340}\) can be further simplified since 340 has a square factor, which is 4 (2²). \(\sqrt{340} = \sqrt{4 \cdot 85} = 2\sqrt{85}\) Therefore, the final simplified answer for cosθ is: cos θ = \(\frac{2\sqrt{85}}{\sqrt{389}}\) And in the answer format requested in the question: 2sqrt(85)/sqrt(389)
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