Finding the period of y=-csc(4x-pi)?
the period is 2pi/b........b is the coefficient of x so 2pi/4 = ?
What is the period of (|sin 4x| +|cos 4x|) ÷ (|sin 4x - cos 4x| + |sin 4x + cos 4x |)?
A2AI havent done questions on finding periods of functions in a while about this but if I were asked to solve this then I would do it this way-(Prerequisites) Period of[math]|\sin[/math][math]\theta|+|[/math][math]\cos[/math][math]\theta[/math]|[math]=\pi/2[/math][math]|\sin\theta+\cos\theta|=\pi[/math]Why?It’s a rough sketch[math]|\sin\theta-\cos\theta|=\pi[/math][math]|\sin\theta+\cos\theta|+|\sin\theta-\cos\theta|=\frac{\pi}{2}[/math]Here’s why[math]\sin 4x= 2\pi/4=\pi/2[/math]Hence,[math]\frac{|\sin 4x|+|\cos 4x|}{|\sin 4x+\cos 4x|+|\sin 4x-\cos 4x|}[/math][math]=\dfrac{\text{Function with period}\:\pi/8}{\text{Function with period}\:\pi/8}[/math]We’ll take LCM[math]=\text{Function with period}\:\pi/8[/math]Like I said before I have some doubts but yeah this is how I’d solve it
What is the period of sin^4x?
The period of the function can be calculated using 2π/|b| where b=4Period: π/2
Find the period of y=2 sin(4x + pi)?
The period would be 2pi/4, which is pi/2. For the general sinusoidal function y = Asin(Bx + C), the period is given by 2pi/B. The frequency is B/(2pi). The amplitude is A, and the phase-shift is C. -John
What is the period of the function y=sin^4x + cos^4x?
The period of this function is (pi/2)When we add two functionsf(x)=g(x)+h(x)If g(x) and h(x) are periodic with fundamental periods T1 and T2. Then the fundamental period of f(x) will beT3=l.c.m(T1,T2)Also if you square a function then the time period will reduce to half of the initial if the function has range including both negative and positive outputs. If only positive or only negative outputs are the range of any function then the time period does not change.Hence here,When sin(x) is squared the time period changes by half. But when sin^2(x) is squared then the time period remains same. Similarly for cos(x) alsoThus for both cos^4(x) and sin^4(x) the fundamental period is (pi). And the lcm of pi and pi is also pi.Hence the time period is (pi)/2.
What are the amplitude, period, and midline of f(x) = -7 sin(4x - π) + 2?
The amplitude is 7 (not -7, as amplitude is always positive). The period is 2Ï€ because it is a sine function and there is no horizontal stretch/compression. The midline is y = 2. (I think - I have never heard the term before). The graph's "new x axis" --- meaning the new line where it is reflected, is y = 2 because the graph is translated up 2 units.
Find the period of the function?
y = a sin b(x±h) ± k a : amplitude => take absolute value b : angular frequency h : horizontal shift ( phase shift). (+ left, - right) k : mid. line or vertical shift ( + up, - down) period = 2π/b ( for sines, cosines. if tangent : p = π/b) 1) y = 1/4 cos 8x b =8 p = 2π/8 = π/4 2) y = 4 sin(−4x) =-4sin(4x) b = 4 p =π/2 3) y = 2 sin(x-pi/3) b = 1 p = 2π 4) y = -8 sin 5(x+pi/5) b = 5 p =2π/5
What is the period of this function, SIN4X+TAN2X?
First of all, [math]\sin(x)[/math] has a period of 2[math]\pi[/math] whereas [math]\tan(x)[/math] has a period of [math]\pi[/math]. You can confirm it from the following plots of the concerned functions:Source: The sin, cos and tan functionsThe next thing is the time periods of the functions [math]\sin(nx)[/math] and [math]\tan(nx)[/math]. The time period of such functions where the parameters (a.k.a angles) are some multiples of [math]x[/math] (not necessarily integral multiples) are calculated as:[math]\frac{\text{Time period of original function}}{n}[/math] ([math]n[/math] being the factor multiplied by the angle [math]x[/math])As an example, consider the function [math]\sin(2\theta)[/math]. This function has the period [math]\frac{2\pi}{2} = \pi[/math]. Similarly, [math]\tan(\frac{1}{2}\theta)[/math] has the period [math]\frac{\pi}{\frac{1}{2}} = 2\pi[/math].Lastly, consider the function [math]\sin(x) + \tan(x)[/math]. What must be it's period? Since [math]\sin(x)[/math] has the period [math]2\pi[/math] and [math]\tan(x)[/math] has the period [math]\pi[/math], so, the sum must have the period [math]2\pi[/math]. Next, consider [math]\sin(2x) + \tan(x)[/math]. Here, both the individual functions have the period [math]\pi[/math]. So, the sum must have a period of [math]\pi[/math]. Now finally, consider the function [math]\sin(2x) + \tan(3x)[/math]. Tell me now, what must be it's period? The former function in the sum repeats every [math]\pi[/math] times, the latter one every [math]\frac{\pi}{3}[/math] times. When will both of them repeat together? At [math]\pi[/math]. So, basically, the function [math]\sin(mx) + \tan(nx)[/math] repeats every [math]LCM(\frac{\pi}{m},\frac{\pi}{n})[/math] times. So, the period of the above function is [math]LCM(\frac{\pi}{m},\frac{\pi}{n})[/math].Coming back to the question,Period of [math]\sin(4x) = \frac{\pi}{2}[/math]Period of [math]\tan(2x) = \frac{\pi}{2}[/math]So, period of [math]\sin(4x) + \tan(2x) = LCM(\frac{\pi}{2}, \frac{\pi}{2})[/math]Which is nothing but [math]\frac{\pi}{2}[/math].Hence, the period of the function [math]\sin(4x) + \tan(2x)[/math] is [math]\frac{\pi}{2}[/math].
Find amplitude, period, and phase shift of y = -1/2sin (4x+3(pi))?
So here are the possible answers to choose from and what I had done. Find amplitude (i), period(ii), and phase shift (iii). y = -1/2sin (4x+3π) A. (i) 1/2 (ii) 4 (iii) -3π/4 B. (i) 2 (ii) π/2 (iii) 3π C. (i) -1/2 (ii) 4 (iii) -4π/3 D. (i) 1/2 (ii) π/2 (iii) -3π/4 _____________________________________ Now, this is what I was taught on solving this: y=Asin(wx-φ) = Asin[w(x-φ/w)] Amplitude: A Period: T=2π/w Phase Shift: φ/w _____________________________________ My work: y=-1/2sin[4(x-3π/4] Amplitude = -1/2 Period: 2π/4 = π/2 Phase Shift: -3π/4 (i) -1/2 (ii) π/2 (iii) -3π/4 So I chose D. (i) 1/2 (ii) π/2 (iii) -3π/4 It is the closest, the only difference being my negative one-half in my answer being a positive in the my multiple choice decision (that was worded poorly, but you smart people get it). :P _____________________________________ I'm taking a test on this tomorrow, I feel confident I am doing this correctly, but I am obviously making some sort of error despite my confidence (unless there is a typo in the review packet). The only thing I think I might be doing wrong is the Phase Shift, as from what I've found online, phi = phase shift. So it could be 3π instead of -3π/4? That still doesn't give me a possible correct multiple choice answer though. Any who, I just want to understand this for future reference: I like knowing what I'm doing. Thank you in advance!