Let #sinx=2/3#, how do you show that #f(2x)=-(4sqrt5)/9#? Trigonometry Right Triangles Relating Trigonometric Functions 1 Answer salamat Mar 14, 2017 see explanation below Explanation: #sin x = 2/3#, therefore by using Pythagoras theorem, #cos x = sqrt 5/3# if #f(x) = sin x#, then #f(2 x) = sin 2 x# #sin 2 x = 2 sin x cos x# #sin 2 x = 2 * 2/3 *sqrt 5/3# therefore, #f (2 x) =sin 2 x = (4 sqrt 5)/9# 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 3323 views around the world You can reuse this answer Creative Commons License