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Show that $\sqrt{\dfrac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\dfrac{1+\tan\alpha}{\tan\alpha-1}$ if $\alpha\in\left(45^\circ;90^\circ\right)$.

We have $\sqrt{\dfrac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\sqrt{\dfrac{\sin^2\alpha+2\sin\alpha\cos\alpha+\cos^2\alpha}{\sin^2\alpha-2\sin\alpha\cos\alpha+\cos^2\alpha}}=\sqrt{\dfrac{(\sin\alpha+\cos\alpha)^2}{(\sin\alpha-\cos\alpha)^2}}=\sqrt{\left(\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\right)^2}.$

Using the fact that $\sqrt{a^2}=|a|$ the given expression is equal to $\left|\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\right|.$ I think that in the inverval $\left(45^\circ;90^\circ\right) \sin\alpha>\cos\alpha$ but how can I prove that? What to do next?

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    $\begingroup$ Given that $\sin(45°)=\cos(45°)$, it is enough to show that the derivative of $\sin(x)$ is larger than the derivative of $\cos(x)$ in $[45°,90°]$. $\endgroup$
    – Anton V.
    Jan 4, 2021 at 8:21
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    $\begingroup$ Divide both sides by $\cos \alpha$. Since it is positive in the given interval, the inequality does not change sign. $\endgroup$
    – Toby Mak
    Jan 4, 2021 at 8:24
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    $\begingroup$ @TobyMak The problem is not the cosine, but the absolute value. $\endgroup$ Jan 4, 2021 at 8:30
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    $\begingroup$ @TobyMak The point of this question is to prove that the denominator is positive. That's why he asks how to prove that $\sin\alpha>\cos\alpha$. And I must say that I'm impressed, if he does not know about derivatives and the variations of trigonometric functions (hence he must be young), but he knows how to write correctly that $\sqrt{a^2}=|a|$. $\endgroup$ Jan 4, 2021 at 8:35
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    $\begingroup$ @TobyMak The OP wants proof of the fact that $\sin x > \cos x$ $x \in (45^{\circ},90^{\circ})$. as the last line of the question suggests. $\endgroup$
    – user822140
    Jan 4, 2021 at 8:43

4 Answers 4

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By the definition of cosine as the x-coordinate of a circle, cosine is negative in the interval $(90^{\circ},180^{\circ})$

$\cos 2x =2\cos^2x-1=1-2\sin^2x$
$\cos 2x =(\sqrt{2} \cos x-1)(\sqrt{2} \cos x+1)=-(\sqrt{2} \sin x-1)(\sqrt{2} \sin x+1)$

Using the fact that $\cos 2x$ is -ve in the interval $x \in (45^{\circ},90^{\circ})$

Prove the fact that $\cos x < \frac{1}{\sqrt{2}}< \sin x$ in the interval $x \in (45^{\circ},90^{\circ})$

Proof 2: There is 1 more way people define $\cos x = \frac{\text{adjacent}}{\text{hypotenuse}}$

Let the angles be $x,90-x,90$

Use the fact that the side opposite to the greater angle is greater.

Therefore, adjacent < opposite (for $x \in (45^{\circ},90^{\circ})$)

Therefore, $\cos x < \sin x$ for $x \in (45^{\circ},90^{\circ})$

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    $\begingroup$ Proof 2 is the most fundamental proof you will find. $\endgroup$
    – user822140
    Jan 4, 2021 at 8:40
  • $\begingroup$ That second proof right there is the best way to do it! +1 $\endgroup$ Jan 4, 2021 at 8:46
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$\sin \alpha + \cos \alpha$ and $\sin \alpha - \cos \alpha$ are both positive in the given domain, so their quotient is also positive, and $|x| = x$ when $x \in \mathbb R^+$. $\sin \alpha + \cos \alpha > 0$ as $\sin \alpha, \cos \alpha$ are positive in the domain. Hence we can focus our attention to just proving $\sin \alpha > \cos \alpha$.

Divide both sides by $\cos \alpha$. Since it is positive in the given interval, the inequality does not change sign.

Thus you have $\tan \alpha > 1$, which is true because $\tan 45º = 1$ and $\tan x$ is a strictly increasing function in the given range $(45º, 90º)$. This is already sufficient as a justification, but on some insight as to why this is, $\tan x = \frac{\sin x}{\cos x}$, and $\sin x$ is increasing while $\cos x$ is decreasing in the given interval, which both increase the value of the function. Anything more rigorous has to involve a geometric argument with the unit circle or calculus.

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Apparently your second question hasn't been answered, about what to do next. Here's what to do: $$\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\equiv\frac{\frac{1}{\cos \alpha}}{\frac{1}{\cos \alpha}}\times\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\equiv\dots$$

I hope that's helpful.

If you need any more help please don't hesitate to ask.

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After your 3rd line, you are actually done, without realizing it; you simply went down the wrong path.

The RHS

$$= \frac{\sin(a) + \cos(a)}{\sin(a) - \cos(a)}.$$

Edit
See the comments following this answer.
A case can be made that my analyis is flawed, since I didn't bother to prove that the denominator above is always positive in the interval $(45^\circ, 90^\circ)$.

Addendum
Responding to

You can prove what the OP wants using pure geometry. Only using the definition of cosine and sine.

Okay: Imagine that you have a unit circle (i.e. of radius $= 1$) centered at the origin, with any point $(x,y)$ that is on the unit circle representing $(\cos[a],\sin[a])$, where $a$ is the angle formed by the two line segments $\overline{(0,0),(1,0)}$ and $\overline{(0,0),(x,y)}$.

Clearly, at $a = 45^\circ$, you have that $(x,y) = \left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right),$ since in a $45^\circ - 45^\circ - 90^\circ$ triangle, the two legs are equal.

Further, you can see geometrically, that for $a$ equal to any angle in $(45^\circ,90^\circ)$, for the corresponding point on the unit circle, the corresponding $y$ coordinate will have increased from $\frac{1}{\sqrt{2}}$, and the corresponding $x$ coordinate will have decreased from $\frac{1}{\sqrt{2}}$.

Therefore, since the $y$ coordinate represents $\sin(a)$ and the $x$ coordinate represents $\cos(a)$, you have that $\sin(a) > \frac{1}{\sqrt{2}} > \cos(a),$ for $a$ equal to any angle in $(45^\circ,90^\circ)$.


A variation on the above argument is to notice that for $a$ equal to any angle in $(45^\circ,90^\circ)$, the slope of the line segment $\overline{(0,0),(x,y)}$ is $> 1$, which (alternatively) implies that $y = \sin(a) > x = \cos(a).$

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    $\begingroup$ Ahem. No. The LHS is always positive, so is the RHS, and your expression is not. $\endgroup$ Jan 4, 2021 at 8:26
  • $\begingroup$ Well, $\sqrt{a^2} = |a|$ not $a$ $\endgroup$
    – user822140
    Jan 4, 2021 at 8:26
  • $\begingroup$ @Jean-ClaudeArbaut "...and your express is not". False, given the constraint that $45^\circ < a < 90^\circ.$ $\endgroup$ Jan 4, 2021 at 8:28
  • $\begingroup$ The point of this question is to prove that. And you didn't. $\endgroup$ Jan 4, 2021 at 8:30
  • $\begingroup$ @Jean-ClaudeArbaut Prove what? It is immediate that if $45^\circ < a < 90^\circ$ then $\sin(a) > \frac{1}{\sqrt{2}} > \cos(a)$. There is nothing to prove. $\endgroup$ Jan 4, 2021 at 8:31

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