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Hey guys,
I am having troubles solving 1A6 (b) in Problem Set 1, Single Variable Calculus. Any ideas?
 one year ago
 one year ago
Hey guys, I am having troubles solving 1A6 (b) in Problem Set 1, Single Variable Calculus. Any ideas?
 one year ago
 one year ago

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dinnertableBest ResponseYou've already chosen the best response.4
Alright so this problem has to do quite a bit with trigonometric identities.To start, write the question as an equation.\[sinx  cosx = Asin(x+c)\]The first thing that comes to mind is that it's possible to rewrite the right side using the compound angle formula.\[sinx  cosx = Asin(x+c)\]\[sinx  cosx = Asinxcosc + Acosxsinc\]From here on it's possible to make a system of equations:\[Asinxcosc = sinx\]\[Acosc = 1\] and\[Acosxsinc = cosx\]\[Asinc = 1\]Now what we're going to do here is make them look like some sort of pythagorean identity so as to simplify them down to one variable.\[Acosc = 1\]\[A^2\cos^2c = 1\]and the other equation,\[Asinc = 1\]\[A^2\sin^sc = 1\]Now the next part may not be quite obvious, but we can solve for A by adding both of the equations together.\[A^2\sin^2c + A^2\cos^2c = 2\]\[A^2(\sin^2c + \cos^2c) = 2\]\[A^2(1) = 2\]\[A=\sqrt{2}\]In the first step, we added together both equations which was in equality to 2. Next we just factored out the A squared so that we could use the pythagorean identity to simplify the equation, and solve for A. So now that we have A solved, we need to solve for c. Rewrite the original equation with the A value that we solved.\[\sqrt{2}sinxcosc + \sqrt{2}cosxsinc = sinx  cosx\]Using the same two system of equations we can now solve for c:\[\sqrt{2}sinxcosc = sinx\]\[\sqrt{2}cosc = 1\]\[cosc = \frac{ 1 }{ \sqrt{2} }\]\[c = \frac{ \pi }{ 4 }, \frac{7\pi}{4}\]Now for the other equation:\[\sqrt{2}cosxsinc = cosx\]\[\sqrt{2}sinc = 1\]\[sinc = \frac{1}{\sqrt{2}}\]\[c = \frac{5\pi}{4}, \frac{7\pi}{4}\]Because c cannot equal two values, we will use the related angles for both equations. \[sinx  cosx = \sqrt{2}sinxcos\frac{7\pi}{4} + \sqrt{2}cosxsin\frac{7\pi}{4}\]\[sinx  cosx = \sqrt{2}\sin(x + \frac{7\pi}{4})\]Although this is right, it is one of the many ways you can express it because sine is periodic. Essentially, this is equivalent to:\[sinx  cosx = \sqrt{2}\sin(x\frac{\pi}{4})\]
 one year ago

dinnertableBest ResponseYou've already chosen the best response.4
Of course once you understand the main concept of the problem, many shortcuts can be made.
 one year ago

dinnertableBest ResponseYou've already chosen the best response.4
Here's a diagram showing the related angles of 7pi/4 and pi/4.
 one year ago

DoobleDBest ResponseYou've already chosen the best response.0
Thank you, understood with your explanation!
 3 months ago
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