How do you calculate #sinh^2(.55)#? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer Shwetank Mauria Oct 11, 2016 #sinh^2(0.55)=0.33426# Explanation: As #sinhx=(e^x-e^(-x))/2#, #sinh(0.55)=(e^0.55-e^(-0.55))/2# = #(1.733253-0.57695)/2# = #1.156303/2# and #sinh^2(0.55)=(1.156303/2)^2# = #0.5781515^2=0.33426# 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 1349 views around the world You can reuse this answer Creative Commons License