What is the value of #costheta# if #sintheta=1/2# and #theta# is in #Q2#? Trigonometry Right Triangles Relating Trigonometric Functions 1 Answer Shwetank Mauria Dec 3, 2016 #costheta=-sqrt3/2# Explanation: As domain of #theta# is #90^o< theta<180^o#, while #sintheta# is positive, #costheta# is negative. As #sintheta=1/2#, and #sin^2theta+cos^2theta=1# we have #cos^2theta=1-sin^2theta# and as #costheta# is negative #costheta=-sqrt(1-sin^2theta)=-sqrt(1-(1/2)^2)# = #-sqrt(1-1/4)=-sqrt(3/4)=-sqrt3/2# Answer link Related questions What does it mean to find the sign of a trigonometric function and how do you find it? What are the reciprocal identities of trigonometric functions? What are the quotient identities for a trigonometric functions? What are the cofunction identities and reflection properties for trigonometric functions? What is the pythagorean identity? If #sec theta = 4#, how do you use the reciprocal identity to find #cos theta#? How do you find the domain and range of sine, cosine, and tangent? What quadrant does #cot 325^@# lie in and what is the sign? How do you use use quotient identities to explain why the tangent and cotangent function have... How do you show that #1+tan^2 theta = sec ^2 theta#? See all questions in Relating Trigonometric Functions Impact of this question 5336 views around the world You can reuse this answer Creative Commons License