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Consider a non­degenerated right triangle with sides of length $\sin x$, $\cos x$, and $\tan x$ where $x$ is a real number. Compute the possible values of the area of this triangle.

  • I was thinking of more along the lines of Heron's formula but that was very nasty indeed, rather I have attempted to find out what are the lengths of the triangle and I have done the Pythagorean theorem which was in vain.
  • Another thing that I have done was to graph all three and to see which one was the largest in the y-value but it turned out that at times one graph was larger than the other and at other times it wasn't.

  • And another thing that I have done was to plug and chug in values such as the number 3 into all of the trig functions and do Pythagorean theorem to see if it satisfied it but none of them didn't

The reason why that I have done those steps was so that I can multiply the legs and divide by two to find the area of the triangle but in order to do that I must know what are the legs.

I was wondering if there was any other way?

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Since $\sin x$, $\cos x$ and $\tan x$ are all positive, we may assume that $\displaystyle x\in\left(0,\frac{\pi}{2}\right)$. Therefore, $\tan x>\sin x$.

If $\cos x>\tan x$, then $\cos x$ is the hypotenuse and

\begin{align} \sin^2x+\tan^2x&=\cos^2x\\ \sin^2x(\cos^2x+1)&=\cos^4x\\ 1-\cos^4x&=\cos^4x\\ \cos x&=\frac{1}{\sqrt[4]{2}} \end{align}

The area of the triangle is

\begin{align} \frac{1}{2}\sin x\tan x&=\frac{\sin^2 x}{2\cos x}\\ &=\frac{1-\cos^2 x}{2\cos x}\\ &=\frac{\displaystyle1-\frac{1}{\sqrt{2}}}{\displaystyle\frac{2}{\sqrt[4]{2}}}\\ &=\frac{\sqrt[4]{2}(2-\sqrt{2})}{4}\\ \end{align}

If $\tan x>\cos x$, then $\tan x$ is the hypotenuse and

\begin{align} \sin^2x+\cos^2x&=\tan^2x\\ x&=\frac{\pi}{4} \end{align}

The area of the triangle is

$$\frac{1}{2}\sin x\cos x=\frac{1}{4}$$

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  • $\begingroup$ For all possible values of x, you have proved there are only two possible values for the area? $\endgroup$ – WGroleau Jun 11 '17 at 11:04
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    $\begingroup$ It is given that $\sin x$, $\cos x$ and $\tan x$ are the three sides of a right-angled triangle. As $\sin x$ cannot be the length of the hypotenuse, there are only two possible cases. In each case, there is only one possible value of $x$. $\endgroup$ – CY Aries Jun 11 '17 at 11:07
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You're told it's a right triangle.

The area of the right triangle is given by $A = \frac 12 ab$, where $a$ and $b$ are the catheti (perpendicular sides).

Since that expression is commutative, you only need to consider the cases $(a,b) = (\sin x, \cos x); (\sin x, \tan x); (\cos x, \tan x)$. You can use the Pythagorean theorem to solve these (knowing the hypotenuse is the remaining trigonometric ratio in each case). One case gives no real solutions for $x$, the other two give valid solutions. Now you should be able to work out the possible expressions for the area.

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  • $\begingroup$ So does that mean that there would be three answers but all written in terms of x? $\endgroup$ – John Rawls Jun 11 '17 at 16:09
  • $\begingroup$ My answer was a heavy hint, not a full solution. When you fully work it out as @CYAries has done, you'll find there are only two numerical ($x$-independent) possible values for the area, $\frac 14$ or $2 - \sqrt 2$. I wanted you to work that out for yourself because it's more fun and rewarding having that final insight. There are only two possibilities, not three, because one of them does not give real solutions for $x$ (or equivalently, as noted in the other answer, no valid right triangle can be formed with those choices). $\endgroup$ – Deepak Jun 11 '17 at 22:43
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tan(x)=+1 or -1 therefore A=(L.W)/2=[cos (x).sin (x)]/2=+1/4 or -1/4 it depends on how the unit vector is oriented.

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