. Find the angles. a) \arctan(1) b) \arctan(-\sqrt{3}) c) \arctan(\frac{1}{\sqrt{3}}) d) \arcsin(-\frac{1}{2}) e) \arcsin(\frac{1}{\sqrt{2}}) f) \arcsin(-\frac{\sqrt{3}}{2}) g) \arccos(\frac{1}{2}) h) \arccos(-\frac{1}{\sqrt{2}}) i) \arccos(\frac{\sqrt{3}}{2}) j) \arcsec(-\sqrt{2}) k) \arcsec(\frac{2}{\sqrt{3}}) l) \arcsec(-2)
The Correct Answer and Explanation is:
To find the angles for each of the inverse trigonometric functions, we can use the standard values and the relationships between trigonometric functions and their inverses. Let’s break down each of the given cases step by step.
a) arctan(1)\arctan(1)arctan(1)
The inverse tangent function, arctan(x)\arctan(x)arctan(x), gives an angle θ\thetaθ such that tan(θ)=x\tan(\theta) = xtan(θ)=x.
For arctan(1)\arctan(1)arctan(1), we need to find the angle where the tangent equals 1. We know that tan(45∘)=1\tan(45^\circ) = 1tan(45∘)=1, so: arctan(1)=45∘orπ4 radians\arctan(1) = 45^\circ \quad \text{or} \quad \frac{\pi}{4} \text{ radians}arctan(1)=45∘or4π radians
b) arctan(−3)\arctan(-\sqrt{3})arctan(−3)
Now we need to find the angle where tan(θ)=−3\tan(\theta) = -\sqrt{3}tan(θ)=−3. We know that tan(60∘)=3\tan(60^\circ) = \sqrt{3}tan(60∘)=3, and for a negative value, we consider the angle in the fourth quadrant: arctan(−3)=−60∘or−π3 radians\arctan(-\sqrt{3}) = -60^\circ \quad \text{or} \quad -\frac{\pi}{3} \text{ radians}arctan(−3)=−60∘or−3π radians
c) arctan(13)\arctan(\frac{1}{\sqrt{3}})arctan(31)
Here, tan(θ)=13\tan(\theta) = \frac{1}{\sqrt{3}}tan(θ)=31, which corresponds to tan(30∘)=13\tan(30^\circ) = \frac{1}{\sqrt{3}}tan(30∘)=31: arctan(13)=30∘orπ6 radians\arctan\left(\frac{1}{\sqrt{3}}\right) = 30^\circ \quad \text{or} \quad \frac{\pi}{6} \text{ radians}arctan(31)=30∘or6π radians
d) arcsin(−12)\arcsin(-\frac{1}{2})arcsin(−21)
The inverse sine function arcsin(x)\arcsin(x)arcsin(x) gives an angle θ\thetaθ such that sin(θ)=x\sin(\theta) = xsin(θ)=x. We are looking for the angle where sin(θ)=−12\sin(\theta) = -\frac{1}{2}sin(θ)=−21. We know that sin(−30∘)=−12\sin(-30^\circ) = -\frac{1}{2}sin(−30∘)=−21: arcsin(−12)=−30∘or−π6 radians\arcsin\left(-\frac{1}{2}\right) = -30^\circ \quad \text{or} \quad -\frac{\pi}{6} \text{ radians}arcsin(−21)=−30∘or−6π radians
e) arcsin(12)\arcsin(\frac{1}{\sqrt{2}})arcsin(21)
For sin(θ)=12\sin(\theta) = \frac{1}{\sqrt{2}}sin(θ)=21, we know sin(45∘)=12\sin(45^\circ) = \frac{1}{\sqrt{2}}sin(45∘)=21: arcsin(12)=45∘orπ4 radians\arcsin\left(\frac{1}{\sqrt{2}}\right) = 45^\circ \quad \text{or} \quad \frac{\pi}{4} \text{ radians}arcsin(21)=45∘or4π radians
f) arcsin(−32)\arcsin(-\frac{\sqrt{3}}{2})arcsin(−23)
For sin(θ)=−32\sin(\theta) = -\frac{\sqrt{3}}{2}sin(θ)=−23, we know that sin(−60∘)=−32\sin(-60^\circ) = -\frac{\sqrt{3}}{2}sin(−60∘)=−23: arcsin(−32)=−60∘or−π3 radians\arcsin\left(-\frac{\sqrt{3}}{2}\right) = -60^\circ \quad \text{or} \quad -\frac{\pi}{3} \text{ radians}arcsin(−23)=−60∘or−3π radians
g) arccos(12)\arccos(\frac{1}{2})arccos(21)
For cos(θ)=12\cos(\theta) = \frac{1}{2}cos(θ)=21, we know cos(60∘)=12\cos(60^\circ) = \frac{1}{2}cos(60∘)=21: arccos(12)=60∘orπ3 radians\arccos\left(\frac{1}{2}\right) = 60^\circ \quad \text{or} \quad \frac{\pi}{3} \text{ radians}arccos(21)=60∘or3π radians
h) arccos(−12)\arccos(-\frac{1}{\sqrt{2}})arccos(−21)
For cos(θ)=−12\cos(\theta) = -\frac{1}{\sqrt{2}}cos(θ)=−21, we know that cos(135∘)=−12\cos(135^\circ) = -\frac{1}{\sqrt{2}}cos(135∘)=−21: arccos(−12)=135∘or3π4 radians\arccos\left(-\frac{1}{\sqrt{2}}\right) = 135^\circ \quad \text{or} \quad \frac{3\pi}{4} \text{ radians}arccos(−21)=135∘or43π radians
i) arccos(32)\arccos(\frac{\sqrt{3}}{2})arccos(23)
For cos(θ)=32\cos(\theta) = \frac{\sqrt{3}}{2}cos(θ)=23, we know that cos(30∘)=32\cos(30^\circ) = \frac{\sqrt{3}}{2}cos(30∘)=23: arccos(32)=30∘orπ6 radians\arccos\left(\frac{\sqrt{3}}{2}\right) = 30^\circ \quad \text{or} \quad \frac{\pi}{6} \text{ radians}arccos(23)=30∘or6π radians
j) \arcsec(−2)\arcsec(-\sqrt{2})\arcsec(−2)
The inverse secant function \arcsec(x)\arcsec(x)\arcsec(x) gives an angle θ\thetaθ such that sec(θ)=x\sec(\theta) = xsec(θ)=x. For sec(θ)=−2\sec(\theta) = -\sqrt{2}sec(θ)=−2, we look for the angle where the secant is −2-\sqrt{2}−2. Since sec(135∘)=−2\sec(135^\circ) = -\sqrt{2}sec(135∘)=−2, we have: \arcsec(−2)=135∘or3π4 radians\arcsec(-\sqrt{2}) = 135^\circ \quad \text{or} \quad \frac{3\pi}{4} \text{ radians}\arcsec(−2)=135∘or43π radians
k) \arcsec(23)\arcsec\left(\frac{2}{\sqrt{3}}\right)\arcsec(32)
For sec(θ)=23\sec(\theta) = \frac{2}{\sqrt{3}}sec(θ)=32, we look for the angle where the secant is 23\frac{2}{\sqrt{3}}32. Since sec(30∘)=23\sec(30^\circ) = \frac{2}{\sqrt{3}}sec(30∘)=32, we have: \arcsec(23)=30∘orπ6 radians\arcsec\left(\frac{2}{\sqrt{3}}\right) = 30^\circ \quad \text{or} \quad \frac{\pi}{6} \text{ radians}\arcsec(32)=30∘or6π radians
l) \arcsec(−2)\arcsec(-2)\arcsec(−2)
For sec(θ)=−2\sec(\theta) = -2sec(θ)=−2, we look for the angle where the secant is −2-2−2. Since sec(120∘)=−2\sec(120^\circ) = -2sec(120∘)=−2, we have: \arcsec(−2)=120∘or2π3 radians\arcsec(-2) = 120^\circ \quad \text{or} \quad \frac{2\pi}{3} \text{ radians}\arcsec(−2)=120∘or32π radians
Conclusion:
The angle values for the given inverse trigonometric functions are as follows:
- a) arctan(1)=45∘\arctan(1) = 45^\circarctan(1)=45∘
- b) arctan(−3)=−60∘\arctan(-\sqrt{3}) = -60^\circarctan(−3)=−60∘
- c) arctan(13)=30∘\arctan(\frac{1}{\sqrt{3}}) = 30^\circarctan(31)=30∘
- d) arcsin(−12)=−30∘\arcsin(-\frac{1}{2}) = -30^\circarcsin(−21)=−30∘
- e) arcsin(12)=45∘\arcsin(\frac{1}{\sqrt{2}}) = 45^\circarcsin(21)=45∘
- f) arcsin(−32)=−60∘\arcsin(-\frac{\sqrt{3}}{2}) = -60^\circarcsin(−23)=−60∘
- g) arccos(12)=60∘\arccos(\frac{1}{2}) = 60^\circarccos(21)=60∘
- h) arccos(−12)=135∘\arccos(-\frac{1}{\sqrt{2}}) = 135^\circarccos(−21)=135∘
- i) arccos(32)=30∘\arccos(\frac{\sqrt{3}}{2}) = 30^\circarccos(23)=30∘
- j) \arcsec(−2)=135∘\arcsec(-\sqrt{2}) = 135^\circ\arcsec(−2)=135∘
- k) \arcsec(23)=30∘\arcsec(\frac{2}{\sqrt{3}}) = 30^\circ\arcsec(32)=30∘
- l) \arcsec(−2)=120∘\arcsec(-2) = 120^\circ\arcsec(−2)=120∘
Each function utilizes basic trigonometric identities and quadrant considerations for accurate angle identification.
