Example Question - inverse trigonometric functions
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Finding the Value of Cosine of Double Arcsine
Claro, vamos a encontrar el valor del coseno de menos dos veces el arco seno de tres quintos. Para esto, utilizaremos algunas identidades trigonométricas. Supongamos \(\theta = \arcsin(\frac{3}{5})\). Entonces, por definición, \(\sin(\theta) = \frac{3}{5}\). Debido a que utilizaremos el coseno de \(-2\theta\), trabajaremos primero encontrando el coseno de \(\theta\) con la ayuda de la identidad fundamental de la trigonometría \(\sin^2(\theta) + \cos^2(\theta) = 1\). \(1. \sin(\theta) = \frac{3}{5}\) \(2. \sin^2(\theta) = (\frac{3}{5})^2\) \(3. \sin^2(\theta) = \frac{9}{25}\) Aplicamos la identidad fundamental de la trigonometría: \(4. \sin^2(\theta) + \cos^2(\theta) = 1\) \(5. \frac{9}{25} + \cos^2(\theta) = 1\) \(6. \cos^2(\theta) = \frac{16}{25}\) \(7. \cos(\theta) = \pm\frac{4}{5}\) Como el arco seno de un valor está entre \(-\frac{\pi}{2}\) y \(\frac{\pi}{2}\), el coseno es positivo. Por lo tanto, \(8. \cos(\theta) = \frac{4}{5}\) Ahora, encontramos el coseno del doble ángulo utilizando la identidad para el coseno de un doble ángulo: \(9. \cos(2\theta) = 2\cos^2(\theta) - 1\) Sustituimos \(\cos(\theta)\) con el valor que calculamos: \(10. \cos(2\theta) = 2(\frac{4}{5})^2 - 1\) \(11. \cos(2\theta) = 2(\frac{16}{25}) - 1\) \(12. \cos(2\theta) = \frac{32}{25} - 1\) \(13. \cos(2\theta) = \frac{32}{25} - \frac{25}{25}\) \(14. \cos(2\theta) = \frac{7}{25}\) Finalmente, como queremos el valor del coseno de menos dos veces \(\theta\), usamos la propiedad de paridad del coseno, que dice que \(\cos(-x) = \cos(x)\). Por tanto, \(15. \cos(-2\theta) = \cos(2\theta)\) \(16. \cos(-2\theta) = \frac{7}{25}\) Así que, el valor de \(\cos(-2\arcsin(\frac{3}{5}))\) es \(\frac{7}{25}\).
Solving Trigonometric Expressions on the Unit Circle
The image shows four trigonometric expressions for which the values need to be found over the interval 0 ≤ θ < 360° using the unit circle. Let's solve each one: 1. sec^(-1)(-√2) = 135° The secant function is the reciprocal of the cosine function. sec(θ) = 1/cos(θ). Since sec(θ) is -√2, we are looking for when cos(θ) = -1/√2. This occurs at an angle of 135°, assuming the principal value (the smallest positive angle). 2. csc^(-1)(2√3/3) = 40° The cosecant function is the reciprocal of the sine function. csc(θ) = 1/sin(θ). Since csc(θ) is 2√3/3, we are looking for when sin(θ) = 3/2√3, which simplifies to 1/√3. This occurs at an angle of 30°. Since cosecant is positive in the first and second quadrants, and the principal value requires the smallest positive angle, the answer would be 30°. 3. cos^(-1)(-√3/2) = 150° The inverse cosine of -√3/2 implies that we are looking for an angle where the cosine value is -√3/2. This happens in the second quadrant, where the cosine is negative, and the corresponding angle is 150°, which is the principal value. 4. sin^(-1)(-1/2) = 210° or 330° The inverse sine of -1/2 implies we are seeking an angle where the sine value is -1/2. This occurs in the third and fourth quadrants where the sine is negative. The corresponding principal angles are 210° and 330°, both of which are solutions over the interval 0 ≤ θ < 360°. However, since we typically take the principal value for the inverse sine, the solution would be the smallest positive angle - 210°. These angles correspond to the values of the angles where the trigonometric functions of the given values are attained on the unit circle within the specified interval.
Solving Quadratic Equations in Terms of tan(x)
The equation provided in the image is a quadratic equation in terms of tan(x): (tan(x))^2 - 1.3tan(x) - 0.48 = 0 To solve it, we can use the quadratic formula: tan(x) = [-b ± sqrt(b^2 - 4ac)] / (2a) Here, a = 1, b = -1.3, and c = -0.48. Plugging these values into the quadratic formula gives us: tan(x) = [1.3 ± sqrt((1.3)^2 - 4(1)(-0.48))] / (2(1)) tan(x) = [1.3 ± sqrt(1.69 + 1.92)] / 2 tan(x) = [1.3 ± sqrt(3.61)] / 2 tan(x) = [1.3 ± 1.9] / 2 This yields two possible solutions for tan(x): tan(x) = (1.3 + 1.9) / 2 = 3.2 / 2 = 1.6 tan(x) = (1.3 - 1.9) / 2 = -0.6 / 2 = -0.3 Now, we will find x for each of these tan(x) values using inverse trigonometric functions and considering the periodicity and symmetry of the tan function, which has a period of π. For tan(x) = 1.6, we can use the arctan function: x1 = arctan(1.6) Because tan(x) is positive in the first and third quadrants, and we're looking for solutions in the interval [0, 2π], we'll also consider: x2 = x1 + π The solutions will be: x1 ≈ arctan(1.6) x2 ≈ arctan(1.6) + π For tan(x) = -0.3, again we use the arctan function: x3 = arctan(-0.3) Since tan(x) is negative in the second and fourth quadrants and within [0, 2π], we will again add π to find an additional solution in the second quadrant: x4 = x3 + π The solutions will be: x3 ≈ arctan(-0.3) x4 ≈ arctan(-0.3) + π You will need to use a calculator to find the numeric values of x1, x3, and add π to find x2 and x4. Make sure that all solutions are within the given interval [0, 2π]. If any solution falls outside this interval, it must be adjusted using the periodicity of the tan function (by adding or subtracting integer multiples of π) to find the equivalent solution within the interval.
Solving Trigonometric Equation Using Quadratic Formula and Inverse Trigonometric Functions
The provided image contains a trigonometric equation and a hint on how to solve it: \[\text{Given: } (\tan(x))^2 - 1.3\tan(x) - 0.48 = 0\] The hint suggests using the quadratic formula, inverse trigonometric functions, and the symmetry of the unit circle to solve the equation. Let's denote \( \tan(x) = y \), which transforms the equation into a quadratic form: \[ y^2 - 1.3y - 0.48 = 0 \] Now, we can solve for \( y \) using the quadratic formula where \( a = 1, b = -1.3, \) and \( c = -0.48 \): \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Applying the values: \[ y = \frac{1.3 \pm \sqrt{(1.3)^2 - 4(1)(-0.48)}}{2} \] Solve under the square root: \[ y = \frac{1.3 \pm \sqrt{1.69 + 1.92}}{2} \] \[ y = \frac{1.3 \pm \sqrt{3.61}}{2} \] \[ y = \frac{1.3 \pm 1.9}{2} \] This gives us two possible solutions for \( y \): \[ y_1 = \frac{1.3 + 1.9}{2} = \frac{3.2}{2} = 1.6 \] \[ y_2 = \frac{1.3 - 1.9}{2} = \frac{-0.6}{2} = -0.3 \] Now, let's turn back to the original variable \( x \): For \( y_1 = \tan(x) = 1.6 \): \[ x = \arctan(1.6) \] This needs to be solved in the given interval \([0, 2\pi]\). Since tangent has a period of \( \pi \), there will be two solutions for \( x \) in the given interval. To find the second solution, we add \( \pi \) to the first solution: \[ x_1 = \arctan(1.6) \] \[ x_2 = \arctan(1.6) + \pi \] For \( y_2 = \tan(x) = -0.3 \): \[ x = \arctan(-0.3) \] Similarly, there will be two solutions for \( x \) in the interval \([0, 2\pi]\): \[ x_3 = \arctan(-0.3) \] \[ x_4 = \arctan(-0.3) + \pi \] Now, let's find the numerical values for these four \( x \) solutions. Remember, these will be approximate values as \( \arctan \) doesn't usually produce neat numbers: \[ x_1 = \arctan(1.6) \] \[ x_2 = \arctan(1.6) + \pi \] \[ x_3 = \arctan(-0.3) \] \[ x_4 = \arctan(-0.3) + \pi \] Using a calculator (or any computational tool): \[ x_1 \approx 1.012197011 \quad(\text{in radians}) \] \[ x_2 \approx 1.012197011 + \pi \approx 4.153789664 \] \[ x_3 \approx -0.291456794 \quad(\text{in radians}) \] \[ x_4 \approx -0.291456794 + \pi \approx 2.850135858 \] Note: \( x_3 \) is negative so we need to add \( 2\pi \) to it to bring it within the \([0, 2\pi]\) interval: \[ x_3 = -0.291456794 + 2\pi \approx 5.992728452 \] The calculator is showing an approximation for the solution \( x \) which looks like: \[ x \approx 0.9828, \arctan(-0.3) + \pi \] We can interpret this as follows: The first solution \( x_1 \) is approximately \( 0.9828 \) (which could be a rounding discrepancy from the more accurate value we computed, \( \approx 1.012197011 \)). The second solution \( x_3 \) is approximately equal to \( \arctan(-0.3) + \pi \), which we computed as \( \approx 2.850135858 \) radians. So, we find \( x \) to have the following approximate solutions in the interval \( [0, 2\pi] \): \[ x_1 \approx 1.012197011 \quad (\text{radians}) \] \[ x_2 \approx 4.153789664 \quad (\text{radians}) \] \[ x_3 \approx 5.992728452 \quad (\text{radians}) \] \[ x_4 \approx 2.850135858 \quad (\text{radians}) \] These solutions correspond to the angles where the original equation holds true within the given interval.
Understanding Inverse Trigonometric Functions
The question in the image asks to evaluate the expression: \[ \cos(\cos^{-1}(-0.9)) \] By definition, the inverse cosine function, \(\cos^{-1}(x)\), returns the angle whose cosine is \(x\). Therefore, when we apply the cosine function to the result of an inverse cosine function, we get back the original value inside the inverse cosine. For the given problem: \[ \cos(\cos^{-1}(-0.9)) = -0.9 \] This is because the inverse cosine of \(-0.9\) gives us the angle whose cosine is \(-0.9\), and taking the cosine of that angle returns us back to the original value of \(-0.9\). So the correct answer is: -0.9
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