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Let's say I want to find the angle measure (in degrees) such that $\tan(x) = -3/2$.

It turns out that $x \approx 123.7$, and when I compute $\tan(123.7)$, I get $\approx -3/2$; so far so good.

However, when I compute $\arctan(-3/2)$, I get $\approx -56.3$ where I expected $\approx 123.7$ since $\arctan$ is the inverse of $\tan$.

jeremy radcliff
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  • How does it turn out that $x \approx 123.7$? How do you know? – Brad Jul 08 '14 at 22:06
  • @Brad It was given in the text, I just don't know how that value was derived though…I thought arctan of the ratio would give the same value, but apparently not. – jeremy radcliff Jul 08 '14 at 22:10

3 Answers3

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By definition, $\arctan x$ is the angle (in radians) between $-\pi/2$ and $\pi/2$ whose tangent is $x$.

There are infinitely many numbers $y$ such that $\tan y=-3/2$.

André Nicolas
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  • But then, how do you find the value of angle x in this case? – jeremy radcliff Jul 08 '14 at 22:01
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    tan(x) is periodic with period $\pi$ or $180^o$ so you should get about 123 by adding 180 to the -56 you get – Jam Jul 08 '14 at 22:02
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    André's point was that you will only get an output of arctan that is within limits of about -1.6 and 1.6 ($-90^o$ and $90^o$)(though there are infinitely many other answers including yours) – Jam Jul 08 '14 at 22:05
  • @oliveeuler Thank you; would you know how that number (123.7) was derived? It's in the context of a rotation transformation in Linear Algebra. It says that the angle of rotation determined by $\tan(x) = -3/2$ is in the second quadrant, does that make a difference? – jeremy radcliff Jul 08 '14 at 22:12
  • If it was given in the text then I have no idea without reading it :) – Jam Jul 08 '14 at 22:25
  • though the quadrants can tell you which value of arctan it is (do you know which angles go to which quadrants?) – Jam Jul 08 '14 at 22:26
  • The "normal" (for angles of a triangle) is between $0$ and $\pi$. In this case, the procedure is simple. If the calculator gives a positive angle, that's the answer. If the calculator (in radian mode) gives a negative number, add $\pi$. In either case, You can if you wish convert to degrees by multiplying by $180/\pi$. I am stressing the radian version, because for mathematics, after a while, it is best that radians become primary. – André Nicolas Jul 08 '14 at 23:17
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I get $−56.3$ where I expected $123.7$ since $\arctan$ is the inverse of $\tan$.

Hint: How much is $180^\circ-56.3^\circ$ ?

Lucian
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The points on the circle corresponding to $-56.3^\circ$ and to $123.7^\circ$ are antipodal to each other, since $-56.3^\circ+ 180^\circ= 123.7^\circ$. The period of the tangent and cotangent functions is only a half circle, i.e. $180^\circ$, rather than a full circle as with the sine, cosine, secant, and cosecant. Thus $\tan(-56.3^\circ)=\tan123.7^\circ$.

So the question is which of these two, $-56.3^\circ$ or $123.7^\circ$, should we take to be the inverse-tangent of $-3/2$? The picture below suggests that taking it to be the negative one allows the inverse tangent function to be continuous, since the tangent does not go up to infinity as the angle stays between $\pm90^\circ$.

Some trigonometric functions

Community
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Michael Hardy
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