tan^1(1)

Question

tan^1(1)

jim_thompson5910 · Answer

Did you mean

$$\Large \tan^{-1}(1)$$

??

anonymous · Answer

i sure did

myininaya · Answer

$$\frac{-\pi}{2}<\tan^{-1}(u)<\frac{\pi}{2} \ \text{ so solve } \tan(\theta)=1 \text{ in that given interval }$$

myininaya · Answer

$$\theta=tan^{-1}(u) \text{ by the way }$$

anonymous · Answer

$$7\pi/4$$

anonymous · Answer

is the answer i keep getting

jim_thompson5910 · Answer

keep in mind that tan(theta) is positive in quadrant 1 and quadrant 3

7pi/4 radians = (7pi/4)*(180/pi) = 315 degrees is in quadrant 4 where tan(theta) is negative

jim_thompson5910 · Answer

also myininaya wrote out the restrictions on arctan: 

$$\large -\frac{\pi}{2}<\tan^{-1}(u)<\frac{\pi}{2}$$

jim_thompson5910 · Answer

because the argument of arctan is positive (positive 1), the restriction is actually reduced to

$$\large 0<\tan^{-1}(u)<\frac{\pi}{2}$$

jim_thompson5910 · Answer

in other words, this angle is in quadrant 1

anonymous · Answer

ok guys sorry i goofed, $$\tan^{-1} (-1)$$

jim_thompson5910 · Answer

then you have the correct answer assuming the book is making the interval [0, 2pi)

but usually the restriction is 

$$\large -\frac{\pi}{2}<\tan^{-1}(u)<\frac{\pi}{2}$$

so that means we have to look for an angle coterminal to 7pi/4 that is in that interval above

jim_thompson5910 · Answer

turns out that -pi/4 and 7pi/4 are coterminal angles, so

tan(-pi/4) = -1
tan(7pi/4) = -1

which is why

$$\large \tan^{-1}(-1) = -\frac{\pi}{4}$$