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Note: Conversion is based on the latest values and formulas.
How do you simplify #sin(x-pi/2)#? - Socratic 16 Aug 2016 · #sin(x- pi/2)# is just #sinx# translated by #(pi/2,0)# We see that #sinx#: graph{sinx [-4.006, 4.006, -2.003, 2.003]} #therefore sin(x-pi/2) #: graph{sin(x- pi/2) [-4.006, 4.006, -2.003, 2.003]} By observing this new graph, we see that this is just #cosx# reflected in the #x# axis. Hence just: #=> -cosx #
How to prove that #sin (pi/2 - theta)# = #cos theta# - Socratic 13 Mar 2016 · using appropriate #color(blue)" Addition formula " # #• sin(A ± B) = sinAcosB ± cosAsinB # hence # sin(pi/2 -theta) = sin(pi/2) costheta - cos(pi/2)sintheta #
How do you evaluate #sin (pi/2)#? - Socratic 15 Aug 2016 · When #theta# becomes equal to #pi/2# then adjacent will vanish and opposite will coincide with the hypotenuse. This means when #theta=pi/2" then # "opposite"="hypotenuse"# So then
How do you verify the identity sin(pi/2 + x) = cosx? | Socratic 23 Apr 2015 · for the "true" proof you need to use matrice, but this is acceptable : sin(a+b) = sin(a)cos(b)+cos(a)sin(b) sin(pi/2+x) = sin(pi/2)*cos(x)+cos(pi/2)*sin(x) sin(pi/2) = 1 cos(pi/2) = 0 So we have : sin(pi/2+x) = cos(x) Since this answer is very usefull for student here the full demonstration to obtain sin(a+b) = sin(a)cos(b)+cos(a)sin(b) (do not read this if you are not fan …
What is #sin(x+pi/2)#? - Socratic 16 Apr 2015 · cos x With pi/2 add to any angle measure, sin changes to cos and vice- versa. Hence It would change to cosine and since the angle measure falls in the second quadrant, hence sin(x+pi/2) would be positive. Alternatively sin(x+pi/2)= sin x cos pi/2 + cos x sinpi/2. Since cos pi/2 is 0 and sinpi/2 is 1, it would be equal to cosx
How do you verify sin(2pi-theta)=-sintheta? - Socratic 8 Mar 2018 · See Below Use the Property: sin(x-y)=sinxcosy-cosxsiny LHS: sin(2pi-theta) =sin2picostheta-cos2pisintheta =0*costheta-1*sintheta =0-sin theta =-sintheta =RHS
How do you simplify # [sin((pi/2)+h)-sin(pi/2)]/(h)#? - Socratic 15 Mar 2016 · How do you apply the fundamental identities to values of #theta# and show that they are true
What is the derivative of #sin((pi/2) - x)#? - Socratic 30 Aug 2016 · -sinx sin(pi/2-x) = cos x, so d/dx(sin(pi/2-x) ) = d/dx(cosx) = -sinx If, for some reason, you don't want to simplify the expression, use d/dx sin(u) = cos u * (du ...
What is the value of Sin^2 (pi/2) - cos (pi)? - Socratic 13 Jul 2016 · sin^2(pi//2)-cos(pi) = 1 - (-1) = 2 To solve this, we need to know the values of the sin and cos functions at specific angles. One of the simplest ways to look at this is using the unit circle.
How do you find the exact values of sin^2(pi/8) using the 28 Jul 2015 · How do you find the exact values of sin^2(pi/8) using the half angle formula? Trigonometry Trigonometric Identities and Equations Half-Angle Identities 1 Answer