Trig Inequality Question (1 Viewer)

rand_althor · Jun 12, 2015

Hey, could someone please help with 2?

$\begin{align*}1&.\textup{ Solve the equation }2\sin^2(x)=1+\cos(2x)\textup{ for }0\leq x \leq 2\pi \\2&. \textup{ Show that if }0<x<\frac{\pi}{4}\textup{, then }\cos(2x)>2\sin^2(x)-1\end{align*}$

From the first question, I can see that there are no points of intersection for the two curves between $0<x<\frac{\pi}{4}$ , but how do I prove the inequality?

InteGrand · Jun 12, 2015

rand_althor said:
Hey, could someone please help with 2?

$\begin{align*}1&.\textup{ Solve the equation }2\sin^2(x)=1+\cos(2x)\textup{ for }0\leq x \leq 2\pi \\2&. \textup{ Show that if }0<x<\frac{\pi}{4}\textup{, then }\cos(2x)>2\sin^2(x)-1\end{align*}$

From the first question, I can see that there are no points of intersection for the two curves between $0<x<\frac{\pi}{4}$ , but how do I prove the inequality?

If there's no points of intersection in that domain, just sub. in a value of x (between $0$ and $\frac{\pi}{4}$ ) into the LHS and and RHS and show the LHS is greater than RHS in this x-value (e.g. use $x=\frac{\pi}{6}$ )). Then since the functions are both continuous on that domain, LHS > RHS for all x in that domain.

rand_althor · Jun 12, 2015

Thanks!

Could you also help with this induction question:

Prove that $n$ distinct lines in a plane which pass through a single given point divide the plane into $2n$ regions.

kawaiipotato · Jun 12, 2015

rand_althor said:
Hey, could someone please help with 2?

$\begin{align*}1&.\textup{ Solve the equation }2\sin^2(x)=1+\cos(2x)\textup{ for }0\leq x \leq 2\pi \\2&. \textup{ Show that if }0<x<\frac{\pi}{4}\textup{, then }\cos(2x)>2\sin^2(x)-1\end{align*}$

From the first question, I can see that there are no points of intersection for the two curves between $0<x<\frac{\pi}{4}$ , but how do I prove the inequality?

1. $1-cos(2x) = 2sin^{2}(x)$
$\therefore 1 - cos(2x) = 1 + cos(2x)$
$cos(2x) = 0$
$$ Then use the fundamental formula for cosine$

2. $Consider, $ cos(2x) - 2sin^{2}(x) + 1$
$= cos(2x) - (1 - cos(2x)) + 1$
$= 2cos2x$
$For 0 < x < \frac{\pi}{4}$
$0 < 2x < \frac{\pi}{2}$
$$ Drawing up a unit circle, $ 1 > cos(2x) > 0$
$2cos(2x) > 0$
$\therefore cos(2x) - 2sin^{2}(x) + 1 > 0$
$cos(2x) > 2sin^{2}(x) - 1$

rand_althor · Jun 12, 2015

rand_althor said:
Thanks!

Could you also help with this induction question:

Prove that $n$ distinct lines in a plane which pass through a single given point divide the plane into $2n$ regions.

Could someone also help with this trig question:

$\textup{Find all angles } \theta\textup{, where }0\leq \theta \leq 2\pi\textup{, for which }\sqrt{2}\sin\theta -\cos\theta=1.$

The answer is: $\theta = \frac{\pi}{3},\pi$

VBN2470 · Jun 12, 2015

rand_althor said:
Could someone also help with this trig question:

$\textup{Find all angles } \theta\textup{, where }0\leq \theta \leq 2\pi\textup{, for which }\sqrt{2}\sin\theta -\cos\theta=1.$

The answer is: $\theta = \frac{\pi}{3},\pi$

Use auxiliary angle method i.e. transformation method to get $$\sqrt{2}\sin\theta -\cos\theta$ = \sqrt{3}\sin(\theta-\arctan{\frac{1}{\sqrt\2}})=1$$ and solve from there.

EDIT: Method I just posted won't help, you probably need to divide out by some trig. function to simplify the equation and solve.

EDIT: Divide out by $$\cos\theta$$ and square both sides to make $$\sec\theta$$ the subject. Everything falls in place after that.

Answer should be $$\theta = \pi, \cos^{-1}\frac{1}{3}$$ .

InteGrand · Jun 12, 2015

rand_althor said:
Thanks!

Could you also help with this induction question:

Prove that $n$ distinct lines in a plane which pass through a single given point divide the plane into $2n$ regions.

Suppose that n such lines divided the plane into 2n regions. For the induction step we just need to show that the next line will introduce two more regions. This is clearly true because the (n+1)^th line will pass through exactly two regions (the spaces between two of the already drawn lines) and divide each of these into two. So in each of those two regions, a new region is made by the (n+1)^th line, so in total, two new regions are formed.

(And the base case of n = 1 is trivial.)

rand_althor · Jun 12, 2015

VBN2470 said:
Use auxiliary angle method i.e. transformation method to get $$\sqrt{2}\sin\theta -\cos\theta$ = \sqrt{3}\sin(\theta-\arctan{\frac{1}{\sqrt\2}})=1$$ and solve from there.

EDIT: Method I just posted won't help, you probably need to divide out by some trig. function to simplify the equation and solve.

EDIT: Divide out by $$\cos\theta$$ and square both sides to make $$\sec\theta$$ the subject. Everything falls in place after that.

Answer should be $$\theta = \pi, \cos^{-1}\frac{1}{3}$$ .

Hey could you please show your working out? I didn't manage to get that answer.

InteGrand · Jun 12, 2015

rand_althor said:
Hey could you please show your working out? I didn't manage to get that answer.

$\sqrt{2}\sin \theta - \cos \theta = 1$

$Divide through by $\cos \theta$: $\sqrt{2}\tan \theta - 1 = \sec \theta$.$

$Square both sides: $2\tan^2 \theta - 2\sqrt{2}\tan \theta + 1 = \underbrace{\sec^2 \theta}_{\tan^2 \theta + 1}$$

$Rearranging, $\tan^2 \theta -2\sqrt{2}\tan \theta = 0$.$

$Now we can easily find solutions. We just need to check them all, because squaring both sides of an equation can introduce spurious solutions.$

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Trig Inequality Question (1 Viewer)

rand_althor

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InteGrand

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rand_althor

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kawaiipotato

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rand_althor

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VBN2470

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InteGrand

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rand_althor

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