Given cos 40=m and sin 10=n,can u please express sin 50 in terms of m and n? Trigonometry Right Triangles Relating Trigonometric Functions 1 Answer Skb · Stefan V. Jul 8, 2017 sin50=sin(40+10)sin50=sin(40+10) =sin40cos10+cos40sin10=sin40cos10+cos40sin10 =sqrt((1-m^2)) * sqrt((1-n^2))+mn=√(1−m2)⋅√(1−n2)+mn Explanation: Use sin(A+B)= sinAcosB+cosAsinBsin(A+B)=sinAcosB+cosAsinB and (cos40)^2+(sin40)^2=1(cos40)2+(sin40)2=1 or (sin40)^2=1-(cos40)^2(sin40)2=1−(cos40)2 or sin40=sqrt([1-(cos40)^2])sin40=√[1−(cos40)2] or sin40=sqrt((1-m^2))sin40=√(1−m2) Similarly, cos40=sqrt((1-n^2))cos40=√(1−n2) 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 = 4secθ=4, how do you use the reciprocal identity to find cos thetacosθ? How do you find the domain and range of sine, cosine, and tangent? What quadrant does cot 325^@cot325∘ 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 theta1+tan2θ=sec2θ? See all questions in Relating Trigonometric Functions Impact of this question 4607 views around the world You can reuse this answer Creative Commons License