How do you evaluate #sin (pi/4) sin (pi/6)#? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer Shwetank Mauria May 3, 2016 #sin(pi/4)xxsin(pi/6)=0.35355# Explanation: #sin(pi/4)xxsin(pi/6)# = #1/sqrt2xx1/2=1/(2sqrt2)=1/4xxsqrt2=1.4142/4=0.35355# Alternatively, #sin(pi/4)xxsin(pi/6)# = #1/2{cos(pi/4-pi/6)-cos(pi/4+pi/6)# = #1/2{cos((3pi-2pi)/12)-cos((3pi+2pi)/12)}# = #1/2{cos(pi/12)-cos((5pi)/12)}# = #1/2{0.9659-0.2588}=1/2xx0.7071=0.35355# 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 3081 views around the world You can reuse this answer Creative Commons License