How do you solve sec (1 2 alpha) = 4? - Socratic 18^@95 + k720^@ 310^@04 + k720^@ sec (x 2) = 1 (sin (x 2)) = 4 sin (x 2) = 1 4 a Calculator gives: sin (x 2) = 1 4 --> x 2 = 14^@48 + k360^@--> x = 28^@95 + k720^@ b
2tan(x)(sin(x)+cos(x))=sec(x)(1+sin2x-cos2x) how to prove this? Since the right hand side is more elaborate, we will modify the right hand side to match that of the left hand side see below 2tanx(sinx+cosx)=secx(1+sin2x-cos2x) Double angle for sin and cos: sin2x= 2sinxcosx cos2x= 1-2sin^2x 2tanx(sinx+cosx)=secx(1+(2sinxcosx)-(1-2sin^2x))= 2tanx(sinx+cosx)=secx(cancel(1)+2sinxcosxcancel(-1)+2sin^2x) Apply the reciprocal identity: secx= 1 cosx 2tanx(sinx
Integration of 1 [ x^2*√(x^2+1)]? - Socratic =-sqrt(x^2+1) x+C We can use trigonometric substitution to solve this Draw a right angle triangle Name one acute angle theta Label the side opposite to it x and the side adjacent to it 1 From Pythagoras' formula which says c^2=a^2+b^2 we know c is the hypotenuse, and a and b are the sides We can write that equation for our triangle: In our case, a is x, b is 1, and c is the hypotenuse