How do you simplify #(sin theta + cos theta) ^2 #? Trigonometry Right Triangles Relating Trigonometric Functions 1 Answer Lovecraft Oct 18, 2015 #(sin(theta) +cos(theta))^2 = sin(2theta) + 1# Explanation: #(sin(theta) +cos(theta))^2 = sin^2(theta) + 2sin(theta)cos(theta) + cos^2(theta)# We know that #sin^2(theta) + cos^2(theta) = 1# so #(sin(theta) +cos(theta))^2 = 2sin(theta)cos(theta) + 1# However, #2sin(theta)cos(theta) = sin(2theta)# so we can say #(sin(theta) +cos(theta))^2 = sin(2theta) + 1# 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 18485 views around the world You can reuse this answer Creative Commons License