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Edit: Although this problem has received a kind answer, I would still appreciate more comprehensive explanation. I am still rather confused.

This problem has confused me a bit:

enter image description here

The standard method is one I know well: Draw a right triangle, set the adjacent and hypotenuse to:$(2x+1)$ and $\sqrt {17}$, respectively; then use the Pythagorean theorem to solve for the third side. Then it is easy to find the sine by taking the opposite over the hypotenuse.

Alas, I feel uncomfortable about this method -- it feels like it's glossing over something. Why can we assume that $\arccos(2x+1/\sqrt {17})$ describes an angle in a RIGHT triangle? The range of $\arccos$ is, after all, zero to $\pi$ -- which obviously includes some values that are too large for a right triangle.

So what gives? Why does the method mentioned above work? Shouldn't we be using unit circle trigonometry rather than the basic right triangle definition?

And, while we're at it, I may as well bring up another similar problem that confused me:

$$\sec(\arctan(\frac{x}{x-1})$$

The teacher said the answer was $\frac{\sqrt{2x^2+1}}{\lvert x-1\rvert}$

Why the absolute value on the denominator? I still can't figure that out. Of course, in a triangle all sides must have positive length, but that doesn't mean we can't have negative values for trig functions!

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    $\begingroup$ I think you are confused: only the angle $\pi/2$ can’t be represented in a right triangle. When we pass into the 2nd quadrant, we use the reference angle to the negative x-axis. $\endgroup$ – H Huang Mar 18 at 16:52
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You need not draw triangles. The relation $$ \lvert\sin\alpha\rvert=\sqrt{1-\cos^2\alpha} $$ holds for every angle $\alpha$ because of $\sin^2\alpha+\cos^2\alpha=1$.

Since $\arccos t$ by definition returns an angle in $[0,\pi]$, you know that $$ \sin(\arccos t)=\sqrt{1-\cos^2\arccos t}=\sqrt{1-t^2} $$ for every $t\in[-1,1]$.

What about $\sec\arctan\frac{x}{x-1}$? You know that $\arctan t$ returns a value in $(-\pi/2,\pi/2)$ where the secant is positive. Thus you can use $$ \lvert\sec\alpha\rvert=\sqrt{1+\tan^2\alpha} $$ and remove the absolute value: $$ \sec\arctan\frac{x}{x-1}=\sqrt{1+\tan^2\arctan\frac{x}{x-1}} =\sqrt{1+\frac{x^2}{(x-1)^2}}=\frac{\sqrt{2x^2-2x+1}}{|x-1|} $$

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  • $\begingroup$ TERRIFIC answer! Thanks!! One small concern: Are you sure the answer you give for the second problem is correct? I ask because my teacher said the answer had root (2x^2+1) in the numerator, rather than root(2x^2-2x+1). $\endgroup$ – Will Mar 20 at 5:26
  • $\begingroup$ Huh... I checked your work and it looks good. I guess my teacher must have mis-graded. $\endgroup$ – Will Mar 20 at 5:35
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When you work with inverse trig functions, it's important to know their ranges. In your first example, we know that $\cos^{-1}$ range is $[0, \pi]$ so the sine value will be non-negative. We do not know whether $2x+1$ is negative or positive but it does not really matter because when we apply Pythagorean theorem, we square $2x+1$, right?. We need to be careful, however, when we take a square root because sine can be negative. Luckily, we know that it's not the case here based on the range of $\arccos$ function.

In your second sample, the range for $\arctan$ is $[-\pi/2, \pi/2]$ so the value of cosine must be non-negative and we need absolute value.

Your observation is correct that the right triangle cannot accommodate negative values for trig functions so I always recommend using unit circle whenever possible.

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  • $\begingroup$ Are you able to expand slightly on this answer? Yours is good as far as it goes, and I was hoping someone else would answer as well and your combined answers would complement one another. However, because no one else has answered (yet) there are still some significant gaps in my understanding. Why, for example, can we assume (2x+1) is positive when creating a right triangle to solve the first problem? If (2x+1) were negative, then we would have to resort to the unit circle - perhaps drawing triangles in multiple quadrants? Something still isn't "clicking" for me here. $\endgroup$ – Will Mar 19 at 15:17
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    $\begingroup$ @Will: I updated my answer. Basically, we do not need to know the sign of the expression because we square it. When we take a square root, that's when we need to look in what quadrant the answer is going to be and decide whether we need a minus before the square root. Simply put, the side length is always positive so you can think about it as an absolute value of trig function. Once you found the absolute value, you need to analyze the sign. $\endgroup$ – Vasya Mar 19 at 15:38
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$$\begin{aligned}\sin\arccos\dfrac{2x+1}{\sqrt{17}}&=\sqrt{1-\left(\dfrac{2x+1}{\sqrt{17}}\right)^2}\\&=\sqrt{\dfrac{17-4x^2-1-4x}{17}}\\&=\dfrac{2}{\sqrt{17}}\sqrt{4-x^2-x}\end{aligned}$$


$$\begin{aligned}\sec\arctan\dfrac{x}{x-1}&=\sqrt{1+\tan^2\arctan\dfrac{x}{x-1}}\\&=\sqrt{1+\left(\dfrac{x}{x-1}\right)^2}\\&=\sqrt{\dfrac{(x-1)^2+x^2}{(x-1)^2}}\\&=\dfrac{\sqrt{2x^2-2x+1}}{\mid x-1\mid}\end{aligned}$$

Clearly range of $\arctan \theta$ is $\left[-\pi/2, \pi/2\right]$. Note that $\arctan\tan\theta \equiv \arctan \sin\theta/\cos\theta$, so $\cos \theta$ must be positive hence the absolute value.

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  • $\begingroup$ Thanks! Did you accidentally switch the plus sign to a minus sign in the first problem, though? $\endgroup$ – Will Mar 20 at 16:54
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    $\begingroup$ Oh @Will Yes I did. I'll correct that. Cheers :)) $\endgroup$ – Paras Khosla Mar 20 at 16:55
  • $\begingroup$ And then the plus 4x needs to change to minus 4x, right? (in the second line) $\endgroup$ – Will Mar 20 at 16:57

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