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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?

Math Student
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    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 $[45°,90°]$. – Anton V. Jan 04 '21 at 08:21
  • Haven't studied that concept. – Math Student Jan 04 '21 at 08:23
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    Divide both sides by $\cos \alpha$. Since it is positive in the given interval, the inequality does not change sign. – Toby Mak Jan 04 '21 at 08:24
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    @TobyMak The problem is not the cosine, but the absolute value. – Jean-Claude Arbaut Jan 04 '21 at 08:30
  • I think the OP already realises that, as $\sin a + \cos a$ is positive in the given domain, but the OP wonders why $\sin a - \cos a$ is also positive. Since the numerator and denominator are both positive in the given domain, then the absolute value sign isn't that big of an issue. I am still waiting for clarification from the OP as they haven't shown that they understand the issue with the absolute value yet. – Toby Mak Jan 04 '21 at 08:32
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    @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 $\sqrt{a^2}=|a|$. – Jean-Claude Arbaut Jan 04 '21 at 08:35
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    @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. –  Jan 04 '21 at 08:43

4 Answers4

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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 $x \in (45^{\circ},90^{\circ})$

Etemon
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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.

Toby Mak
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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.

A-Level Student
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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 $y = \sin(a) > x = \cos(a).$

user2661923
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    Ahem. No. The LHS is always positive, so is the RHS, and your expression is not. – Jean-Claude Arbaut Jan 04 '21 at 08:26
  • Well, $\sqrt{a^2} = |a|$ not $a$ –  Jan 04 '21 at 08:26
  • @Jean-ClaudeArbaut "...and your express is not". False, given the constraint that $45^\circ < a < 90^\circ.$ – user2661923 Jan 04 '21 at 08:28
  • The point of this question is to prove that. And you didn't. – Jean-Claude Arbaut Jan 04 '21 at 08:30
  • @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. – user2661923 Jan 04 '21 at 08:31
  • Well, no, it's not necessarily immediate. It all depends on what you know about trigonometric functions. And if the OP does not know about derivatives, he has to do it geometrically. Not very difficult, but not immediate, and anyway you didn't address this. – Jean-Claude Arbaut Jan 04 '21 at 08:33
  • @Jean-ClaudeArbaut Now that I think of it, you have a point. However, I still question this. Typically, a pre-calculus student will be familiar with the graphs of $\sin(x)$ and $\cos(x)$ which intersect at $45^\circ$, with $\sin(x)$ increasing and $\cos(x)$ decreasing. However, as I say, it is plausible that that may have been part of the point of the problem, and you are right, I just assumed that it was common knowledge, re familiarity with the graphs. – user2661923 Jan 04 '21 at 08:37