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lucidriled

  • 4 years ago

Hi, for part A of differentiation problem set, i don't understand problems 6B and 7B for 1A. 6B) asks to state "sinx-cosx" in "Asin (x+c)" form, with the answer being "√2 sin(x −π/4)" 7B) asks for the period , amplitude , and phase angle, to sketch "−4cos(x + π/2)." How does that equal "4sin x" ? Any clarification would be very much appreciated?

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  1. lucidriled
    • 4 years ago
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    oh, just to be clear that last sentence should not be a question

  2. ac7qz
    • 4 years ago
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    Hi, for problem 6b, we use the formula \[\sin (a+b) = \sin a \cos b + \cos a \sin b\], then fit it to the formula \[f(x) = \sin x - \cos x=A \sin x \cos c -A \cos x \sin c\]. Equating the sin x terms we get A cos c = 1, and from the cos x terms we get A sin c = -1. If cos c = - sin c, then c must be - pi/4. Substituting back in either equation to get A gives us sqrt 2. (Please excuse the notation, I'm unfamiliar with the equation editor.) For problem 7b, we have a - cosine with an amplitude of 4 and translated pi/4 to the left. That is equivalent to + sin. remember every pi/4 translation changes between sine and cosine, and every pi/2 translation just switches the sign of the coefficient.

  3. lucidriled
    • 4 years ago
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    Hi, thanks for the response. I would have responded sooner but open study does not send email notifications apparently. Let me try to explain the answers to make sure I understand them properly. For question 6B: We recognize that sinx - cosx fits into the basic trig identity where sin (x-c) equals Asinxcosc - Acosxsinc, with A being any constant that is not necessarily needed in the equation. In order for the later to equal to sinx - cosx, Acosc must equal 1 and Asinc must equal -1. Using the unit circle, we find that where cosc=-sinc it is -pi/4 or 7pi/4. Do we use the negative one because sin (x-c) contains a negative c? I don't get exactly how A= sqrt (2). Every time I try to solve for it, I have to perform an illegal operation to get the answer like subtracting instead of dividing. For 7B: The negative sign inverts cos and by translating it by (pi/2) you align the function with the sin function. You wrote pi/4? Was that intended or should it be pi/2? I've heard of the sin,cos, and the sign of coefficient change with translation but I don't exactly get the concept. I get that if you move cos or sin function over by pi/2 it will align with the other, so we can now say it's function behaves in the same way as the other. I don't know what I'm checking for to see the change in sign of the coefficient when it moves pi/4. It seems that the slopes remain the same. I apologize if the questions and explanations seem elementary, but it has been quite some time since I done any math at all. Again, any clarification would be very much appreciated.

  4. Saber_Shakibi
    • 4 years ago
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    Hi, For Question 6B: We can use \[2k \pi \pm -\pi/4\] k being any integer to be general but for simplicity we use one of them so others are valid too. and since we know that \[\sin( 2k \pi \pm -\pi/4)= -\sqrt{2}/2=-1/\sqrt{2}\] and \[\cos (2k \pi \pm -\pi/4)= \sqrt{2}/2=1/\sqrt{2}\] we multiply whole quantity by \[\sqrt{2}\] to get our 1 and -1 back. For Question 7B: we have some formulas for changing the angle of sin and cos: \[\sin(x+\pi/2)=\cos x\] \[\sin(x-\pi/2)=-\cos x\] \[\sin(\pi/2-x)=\cos x\] \[\sin (x \pm \pi)=- \sin x\] \[\cos(x+\pi/2)=-\sin x\] \[\cos (x-\pi/2)= \sin x\] \[\cos(\pi/2-x)=\sin x\] \[\cos(x \pm \pi)=-\cos x\] all of them can be proofed by the two equations below: \[\sin (a+b)= \sin a \times \cos b +\sin b \times \cos a\] \[\cos (a+b)=\cos a \times \cos b-\sin a \times \sin b\] I hope it'll help you.

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