<p>Findthe values of x on the interval [0,π] for which cos x < sin 2x. (WITHOUT using a graphic calculator). </p>
<p>I found it in the Barron's book but they only give teach you how to solve it using a graphic calculator. Can any one teach me the traditional way. It'd be very helpful. </p>
<p>Definitely possible if you know some slightly more advanced trig.</p>
<p>First, do a very quick rough sketch of sin2x and cosx for 0<x<π. It should take just take a couple of seconds (one full period of sin, half of cos). From this you can see:</p>
<p>-The graphs touch at 3 points, somewhere between 0 and π/2, at π/2, and somewhere between π/2 and π.
-The answer you need is: [Point 1 < x < π/2 AND Point 2 < x < π].</p>
<p>Now to find the points of intersection:</p>
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<p>You might know from memory that sin(π/6) = 1/2 or you could get that by putting arcsin(1/2) into your calc. There are several ways to get the other value. I used the general solution:</p>
<p>For sin(θ)=sin(α), the general solution is : θ = nπ+ [(−1)^(n)]*α (last term is [minus one to the power n], times alpha)</p>
<p>Basically, this formula gives angles θ for n=1,2,3,4,5… such that sin(θ) = sin(α) where α is whatever you want it to be.</p>
<p>n=0 =>
θ = 0 + 1*α = α. [Minus one to the power zero is one].
So the first angle whose sine = sin(α) is α (Duh!). </p>
<p>n=1 => </p>
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<p>Therefore the answer is [π/6 < x < π/2] and [5π/6 < x < π] ------------------------------------</p>
<p>All in all, you could do that calculation in around a minute or less if you’re familiar with the general solution formula (looks a lot more complicated than it is). Or you might already know that sin(π - α) = sin(α) in which case you could bypass it altogether.</p>
<p>My solution is similar to Iostint’s but we divide both sides of the inequality cos x < 2 sin x cos x by cos x, noting that the inequality sign changes if cos x < 0.</p>
<p>By double-angle and dividing by cos x, cos x < 2 sin x cos x iff 1 < 2 sin x and cos x > 0, OR 1 > 2 sin x and cos x < 0. (note that cos x = 0 implies 0 < 0, which is not true)</p>
<p>If 1 < 2 sin x and cos x > 0, then sin x > 1/2, cos x > 0 and x is in the interval (pi/6, pi/2).</p>
<p>If 1 > 2 sin x and cos x < 0, then sin x < 1/2, cos x < 0, and x is in the interval (5pi/6, pi) (x can be in (pi, 3pi/2) but the domain of x is restricted to [0, pi]).</p>
<p>Answer = (pi/6, pi/2) U (5pi/6, pi). Note that the endpoints are <em>not</em> included.</p>
<p>D’oh. Typical me, making a mountain out of a molehill. I just typed up the first method I could think of, but your method is much better/quicker.</p>