Question #b6b30 Trigonometry Right Triangles Relating Trigonometric Functions 1 Answer P dilip_k Feb 21, 2017 Given #sinx=1/4 # # and 0< x < pi/2-> x in "1st quadrant"# So #cosx=sqrt(1-sin^2x)=sqrt(1-1/16)=sqrt15/4# Now #sin(x/2)=sqrt(1/2(1-cosx))# #=sqrt(1/2(1-sqrt15/4))# #=sqrt(1/8(4-sqrt15)# 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 1397 views around the world You can reuse this answer Creative Commons License