How do I find the value of sin 105? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer Konstantinos Michailidis Sep 24, 2015 It is #sin105^o=1/4*(sqrt6+sqrt2)# Explanation: We have that #sin(105)=sin(60+45)# because #sin(a+b)=sina*cosb+cosa*sinb# hence, we have #sin105=sin(60+45)=sin60*cos45+cos60*sin45= =sqrt3/2*sqrt2/2+1/2*sqrt2/2=1/4(sqrt6+sqrt2)# Answer link Related questions How do you find the trigonometric functions of any angle? What is the reference angle? How do you use the ordered pairs on a unit circle to evaluate a trigonometric function of any angle? What is the reference angle for #140^\circ#? How do you find the value of #cot 300^@#? What is the value of #sin -45^@#? How do you find the trigonometric functions of values that are greater than #360^@#? How do you use the reference angles to find #sin210cos330-tan 135#? How do you know if #sin 30 = sin 150#? How do you show that #(costheta)(sectheta) = 1# if #theta=pi/4#? See all questions in Trigonometric Functions of Any Angle Impact of this question 31399 views around the world You can reuse this answer Creative Commons License