{"id":44491,"date":"2025-06-30T16:48:34","date_gmt":"2025-06-30T16:48:34","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=44491"},"modified":"2025-06-30T16:48:35","modified_gmt":"2025-06-30T16:48:35","slug":"find-the-angles","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/find-the-angles\/","title":{"rendered":"Find the angles."},"content":{"rendered":"\n

. 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)<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

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\u2019s break down each of the given cases step by step.<\/p>\n\n\n\n

a) arctan\u2061(1)\\arctan(1)arctan(1)<\/h3>\n\n\n\n

The inverse tangent function, arctan\u2061(x)\\arctan(x)arctan(x), gives an angle \u03b8\\theta\u03b8 such that tan\u2061(\u03b8)=x\\tan(\\theta) = xtan(\u03b8)=x.
For arctan\u2061(1)\\arctan(1)arctan(1), we need to find the angle where the tangent equals 1. We know that tan\u2061(45\u2218)=1\\tan(45^\\circ) = 1tan(45\u2218)=1, so: arctan\u2061(1)=45\u2218or\u03c04 radians\\arctan(1) = 45^\\circ \\quad \\text{or} \\quad \\frac{\\pi}{4} \\text{ radians}arctan(1)=45\u2218or4\u03c0\u200b radians<\/p>\n\n\n\n

b) arctan\u2061(\u22123)\\arctan(-\\sqrt{3})arctan(\u22123\u200b)<\/h3>\n\n\n\n

Now we need to find the angle where tan\u2061(\u03b8)=\u22123\\tan(\\theta) = -\\sqrt{3}tan(\u03b8)=\u22123\u200b. We know that tan\u2061(60\u2218)=3\\tan(60^\\circ) = \\sqrt{3}tan(60\u2218)=3\u200b, and for a negative value, we consider the angle in the fourth quadrant: arctan\u2061(\u22123)=\u221260\u2218or\u2212\u03c03 radians\\arctan(-\\sqrt{3}) = -60^\\circ \\quad \\text{or} \\quad -\\frac{\\pi}{3} \\text{ radians}arctan(\u22123\u200b)=\u221260\u2218or\u22123\u03c0\u200b radians<\/p>\n\n\n\n

c) arctan\u2061(13)\\arctan(\\frac{1}{\\sqrt{3}})arctan(3\u200b1\u200b)<\/h3>\n\n\n\n

Here, tan\u2061(\u03b8)=13\\tan(\\theta) = \\frac{1}{\\sqrt{3}}tan(\u03b8)=3\u200b1\u200b, which corresponds to tan\u2061(30\u2218)=13\\tan(30^\\circ) = \\frac{1}{\\sqrt{3}}tan(30\u2218)=3\u200b1\u200b: arctan\u2061(13)=30\u2218or\u03c06 radians\\arctan\\left(\\frac{1}{\\sqrt{3}}\\right) = 30^\\circ \\quad \\text{or} \\quad \\frac{\\pi}{6} \\text{ radians}arctan(3\u200b1\u200b)=30\u2218or6\u03c0\u200b radians<\/p>\n\n\n\n

d) arcsin\u2061(\u221212)\\arcsin(-\\frac{1}{2})arcsin(\u221221\u200b)<\/h3>\n\n\n\n

The inverse sine function arcsin\u2061(x)\\arcsin(x)arcsin(x) gives an angle \u03b8\\theta\u03b8 such that sin\u2061(\u03b8)=x\\sin(\\theta) = xsin(\u03b8)=x. We are looking for the angle where sin\u2061(\u03b8)=\u221212\\sin(\\theta) = -\\frac{1}{2}sin(\u03b8)=\u221221\u200b. We know that sin\u2061(\u221230\u2218)=\u221212\\sin(-30^\\circ) = -\\frac{1}{2}sin(\u221230\u2218)=\u221221\u200b: arcsin\u2061(\u221212)=\u221230\u2218or\u2212\u03c06 radians\\arcsin\\left(-\\frac{1}{2}\\right) = -30^\\circ \\quad \\text{or} \\quad -\\frac{\\pi}{6} \\text{ radians}arcsin(\u221221\u200b)=\u221230\u2218or\u22126\u03c0\u200b radians<\/p>\n\n\n\n

e) arcsin\u2061(12)\\arcsin(\\frac{1}{\\sqrt{2}})arcsin(2\u200b1\u200b)<\/h3>\n\n\n\n

For sin\u2061(\u03b8)=12\\sin(\\theta) = \\frac{1}{\\sqrt{2}}sin(\u03b8)=2\u200b1\u200b, we know sin\u2061(45\u2218)=12\\sin(45^\\circ) = \\frac{1}{\\sqrt{2}}sin(45\u2218)=2\u200b1\u200b: arcsin\u2061(12)=45\u2218or\u03c04 radians\\arcsin\\left(\\frac{1}{\\sqrt{2}}\\right) = 45^\\circ \\quad \\text{or} \\quad \\frac{\\pi}{4} \\text{ radians}arcsin(2\u200b1\u200b)=45\u2218or4\u03c0\u200b radians<\/p>\n\n\n\n

f) arcsin\u2061(\u221232)\\arcsin(-\\frac{\\sqrt{3}}{2})arcsin(\u221223\u200b\u200b)<\/h3>\n\n\n\n

For sin\u2061(\u03b8)=\u221232\\sin(\\theta) = -\\frac{\\sqrt{3}}{2}sin(\u03b8)=\u221223\u200b\u200b, we know that sin\u2061(\u221260\u2218)=\u221232\\sin(-60^\\circ) = -\\frac{\\sqrt{3}}{2}sin(\u221260\u2218)=\u221223\u200b\u200b: arcsin\u2061(\u221232)=\u221260\u2218or\u2212\u03c03 radians\\arcsin\\left(-\\frac{\\sqrt{3}}{2}\\right) = -60^\\circ \\quad \\text{or} \\quad -\\frac{\\pi}{3} \\text{ radians}arcsin(\u221223\u200b\u200b)=\u221260\u2218or\u22123\u03c0\u200b radians<\/p>\n\n\n\n

g) arccos\u2061(12)\\arccos(\\frac{1}{2})arccos(21\u200b)<\/h3>\n\n\n\n

For cos\u2061(\u03b8)=12\\cos(\\theta) = \\frac{1}{2}cos(\u03b8)=21\u200b, we know cos\u2061(60\u2218)=12\\cos(60^\\circ) = \\frac{1}{2}cos(60\u2218)=21\u200b: arccos\u2061(12)=60\u2218or\u03c03 radians\\arccos\\left(\\frac{1}{2}\\right) = 60^\\circ \\quad \\text{or} \\quad \\frac{\\pi}{3} \\text{ radians}arccos(21\u200b)=60\u2218or3\u03c0\u200b radians<\/p>\n\n\n\n

h) arccos\u2061(\u221212)\\arccos(-\\frac{1}{\\sqrt{2}})arccos(\u22122\u200b1\u200b)<\/h3>\n\n\n\n

For cos\u2061(\u03b8)=\u221212\\cos(\\theta) = -\\frac{1}{\\sqrt{2}}cos(\u03b8)=\u22122\u200b1\u200b, we know that cos\u2061(135\u2218)=\u221212\\cos(135^\\circ) = -\\frac{1}{\\sqrt{2}}cos(135\u2218)=\u22122\u200b1\u200b: arccos\u2061(\u221212)=135\u2218or3\u03c04 radians\\arccos\\left(-\\frac{1}{\\sqrt{2}}\\right) = 135^\\circ \\quad \\text{or} \\quad \\frac{3\\pi}{4} \\text{ radians}arccos(\u22122\u200b1\u200b)=135\u2218or43\u03c0\u200b radians<\/p>\n\n\n\n

i) arccos\u2061(32)\\arccos(\\frac{\\sqrt{3}}{2})arccos(23\u200b\u200b)<\/h3>\n\n\n\n

For cos\u2061(\u03b8)=32\\cos(\\theta) = \\frac{\\sqrt{3}}{2}cos(\u03b8)=23\u200b\u200b, we know that cos\u2061(30\u2218)=32\\cos(30^\\circ) = \\frac{\\sqrt{3}}{2}cos(30\u2218)=23\u200b\u200b: arccos\u2061(32)=30\u2218or\u03c06 radians\\arccos\\left(\\frac{\\sqrt{3}}{2}\\right) = 30^\\circ \\quad \\text{or} \\quad \\frac{\\pi}{6} \\text{ radians}arccos(23\u200b\u200b)=30\u2218or6\u03c0\u200b radians<\/p>\n\n\n\n

j) \\arcsec(\u22122)\\arcsec(-\\sqrt{2})\\arcsec(\u22122\u200b)<\/h3>\n\n\n\n

The inverse secant function \\arcsec(x)\\arcsec(x)\\arcsec(x) gives an angle \u03b8\\theta\u03b8 such that sec\u2061(\u03b8)=x\\sec(\\theta) = xsec(\u03b8)=x. For sec\u2061(\u03b8)=\u22122\\sec(\\theta) = -\\sqrt{2}sec(\u03b8)=\u22122\u200b, we look for the angle where the secant is \u22122-\\sqrt{2}\u22122\u200b. Since sec\u2061(135\u2218)=\u22122\\sec(135^\\circ) = -\\sqrt{2}sec(135\u2218)=\u22122\u200b, we have: \\arcsec(\u22122)=135\u2218or3\u03c04 radians\\arcsec(-\\sqrt{2}) = 135^\\circ \\quad \\text{or} \\quad \\frac{3\\pi}{4} \\text{ radians}\\arcsec(\u22122\u200b)=135\u2218or43\u03c0\u200b radians<\/p>\n\n\n\n

k) \\arcsec(23)\\arcsec\\left(\\frac{2}{\\sqrt{3}}\\right)\\arcsec(3\u200b2\u200b)<\/h3>\n\n\n\n

For sec\u2061(\u03b8)=23\\sec(\\theta) = \\frac{2}{\\sqrt{3}}sec(\u03b8)=3\u200b2\u200b, we look for the angle where the secant is 23\\frac{2}{\\sqrt{3}}3\u200b2\u200b. Since sec\u2061(30\u2218)=23\\sec(30^\\circ) = \\frac{2}{\\sqrt{3}}sec(30\u2218)=3\u200b2\u200b, we have: \\arcsec(23)=30\u2218or\u03c06 radians\\arcsec\\left(\\frac{2}{\\sqrt{3}}\\right) = 30^\\circ \\quad \\text{or} \\quad \\frac{\\pi}{6} \\text{ radians}\\arcsec(3\u200b2\u200b)=30\u2218or6\u03c0\u200b radians<\/p>\n\n\n\n

l) \\arcsec(\u22122)\\arcsec(-2)\\arcsec(\u22122)<\/h3>\n\n\n\n

For sec\u2061(\u03b8)=\u22122\\sec(\\theta) = -2sec(\u03b8)=\u22122, we look for the angle where the secant is \u22122-2\u22122. Since sec\u2061(120\u2218)=\u22122\\sec(120^\\circ) = -2sec(120\u2218)=\u22122, we have: \\arcsec(\u22122)=120\u2218or2\u03c03 radians\\arcsec(-2) = 120^\\circ \\quad \\text{or} \\quad \\frac{2\\pi}{3} \\text{ radians}\\arcsec(\u22122)=120\u2218or32\u03c0\u200b radians<\/p>\n\n\n\n

Conclusion:<\/h3>\n\n\n\n

The angle values for the given inverse trigonometric functions are as follows:<\/p>\n\n\n\n

    \n
  • a) arctan\u2061(1)=45\u2218\\arctan(1) = 45^\\circarctan(1)=45\u2218<\/li>\n\n\n\n
  • b) arctan\u2061(\u22123)=\u221260\u2218\\arctan(-\\sqrt{3}) = -60^\\circarctan(\u22123\u200b)=\u221260\u2218<\/li>\n\n\n\n
  • c) arctan\u2061(13)=30\u2218\\arctan(\\frac{1}{\\sqrt{3}}) = 30^\\circarctan(3\u200b1\u200b)=30\u2218<\/li>\n\n\n\n
  • d) arcsin\u2061(\u221212)=\u221230\u2218\\arcsin(-\\frac{1}{2}) = -30^\\circarcsin(\u221221\u200b)=\u221230\u2218<\/li>\n\n\n\n
  • e) arcsin\u2061(12)=45\u2218\\arcsin(\\frac{1}{\\sqrt{2}}) = 45^\\circarcsin(2\u200b1\u200b)=45\u2218<\/li>\n\n\n\n
  • f) arcsin\u2061(\u221232)=\u221260\u2218\\arcsin(-\\frac{\\sqrt{3}}{2}) = -60^\\circarcsin(\u221223\u200b\u200b)=\u221260\u2218<\/li>\n\n\n\n
  • g) arccos\u2061(12)=60\u2218\\arccos(\\frac{1}{2}) = 60^\\circarccos(21\u200b)=60\u2218<\/li>\n\n\n\n
  • h) arccos\u2061(\u221212)=135\u2218\\arccos(-\\frac{1}{\\sqrt{2}}) = 135^\\circarccos(\u22122\u200b1\u200b)=135\u2218<\/li>\n\n\n\n
  • i) arccos\u2061(32)=30\u2218\\arccos(\\frac{\\sqrt{3}}{2}) = 30^\\circarccos(23\u200b\u200b)=30\u2218<\/li>\n\n\n\n
  • j) \\arcsec(\u22122)=135\u2218\\arcsec(-\\sqrt{2}) = 135^\\circ\\arcsec(\u22122\u200b)=135\u2218<\/li>\n\n\n\n
  • k) \\arcsec(23)=30\u2218\\arcsec(\\frac{2}{\\sqrt{3}}) = 30^\\circ\\arcsec(3\u200b2\u200b)=30\u2218<\/li>\n\n\n\n
  • l) \\arcsec(\u22122)=120\u2218\\arcsec(-2) = 120^\\circ\\arcsec(\u22122)=120\u2218<\/li>\n<\/ul>\n\n\n\n

    Each function utilizes basic trigonometric identities and quadrant considerations for accurate angle identification.<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    . 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 […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-44491","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/44491","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=44491"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/44491\/revisions"}],"predecessor-version":[{"id":44493,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/44491\/revisions\/44493"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=44491"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=44491"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=44491"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}