How do you get cos2t = 1-2sin^2(t)?

Question

I have to work out how to go from the expression

cos(a+b) = cos(a)cos(b) -sin(a)sin(b)

Can anyone tell me the next step so I can work it out?

to prove that

cos2t = 1-2sin^2(t)

Answer 1

in your equation ,let a = b = t
hence substitute in your equation
R.H.S. : cos( a + b) = cos( t + t) = cos 2t
L.H.S. : cos(t) cos( t) - sin (t) sin (t)
.............= {cos (t) }^2 - { sin(t) }^2
.............= 1 - {sin (t)}^2 - {sin(t)}^2 ... [ ( cos x )^2 = 1 - (sin x)^2]
.............= 1 - 2{ sin(t)}^2

hence proved

Answer 2

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RE:
How do you get cos2t = 1-2sin^2(t)?
I have to work out how to go from the expression

cos(a+b) = cos(a)cos(b) -sin(a)sin(b)

Can anyone tell me the next step so I can work it out?

to prove that

cos2t = 1-2sin^2(t)

Answer 3

You need this equation sin^2 a + cos^2 a = 1
. . . . or cos^2 a = 1 - sin^2 a . . . . equation 1
from your equation
cos ( a + b ) = cos a cos b - sin a sin b . . . . .let a = b . . then
cos ( 2 a ) = cos a cos a - sin a sin a
. . . . . . = cos^2 a - sin^2 a . . . substitute equation 1
. . . . . . = 1 - sin^2 a - sin^2 a
. . . . . . = 1 - 2 sin^2 a . . . . just change the variable

cos 2t = 1 - 2 sin^2 t

Answer 4

Cos 2t

Answer 5

cos(2t)=cos(t+t)= costxcost-sintxsint=cos^2t-sin^2 t
=1-sin^t -sin^2 t=1-2sin^2 t QED

Answer 6

cos(a+b) = {cos (a).cos(b) - sin(a).sin(b)}

let a=t and b=t

cos(t+t) = { cos(t).cos(t) - sin(t).sin(t)}

cos(2t) = cos^2(t) - sin^2(t)

BUT cos^2(t)=1-sin^2(t)
therefore
cos(2t) = 1-sin^2(t)-sin^2(t)

cos(2t) = 1-2sin^2(t)

Answer 7

if a=b then we get cos(a+a)=cosa.cosa-sina.sina=cos2a-sin2a=cos2a

thereforeif a=t weget cos 2t=cos2a-sin2a therefore cos 2a=cos2a-sin2a=1-sin2a-sin2a(ascos2a+sin2a=1) =1-2sin2t

Answer 8

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Given x=csin2t(1+cos2t) and y=ccos2t(1-cos2t) Using uv rule, dx/dt = csin2t (0-2sin2t) + c(1+cos2t) . (2cos2t) = -2c sin^2 2t + 2c cos2t + 2c cos^2 2t Combining first and last terms,we get = 2c (cos^2 2t - sin^2 2t) + 2c cos2t = 2c(cos4t) + 2c cos2t (since cos^2 x - sin^2 x = cos2x) = 2c(cos4t + cos2t) = 2c(2cos3t . cost) [ since cosa +cosb = cos( (a+b)/2 ) . cos( (a-b)/2) ) ] dx/dt = 4c cos3t . cost ---------------(1) Using uv rule, dy/dt = c cos2t (0+2sin2t) + c(1-cos2t) . (-2sin2t) = 2c sin2t . cos2t - 2c sin2t + 2c sin2t . cos2t Combining first and last terms, we get = 2 (2c sin2t . cos2t) - 2c sin2t [ since 2sinx . cosx = sin2x ] = 2 (c sin4t) - 2c sin2t = 2c (sin4t - sin2t) = 2c (2cos3t . sint) [ since sina - sinb = 2cos( (a+2)/2 ) . sin( (a-2)/2 ) ] dy/dt = 4c cos3t . sint ---------------(2) Now (2) / (1), we get dy/dx = (4c cos3t . sint) / (4c cos3t . cost) = tant => y1 = tant ---------------------(3) Differentiating (3) w.r.t. x we get, y2 = sec^2 t .(dx/dt) = sec^2 t .[ 1/ (4c cos3t . cost)] y2 = (sec^3 t) / (4c cos3t) --------------------(4) Now, radius of curvature = [ ( {1+(y1)^2}^(3/2) ) / y2] = [ (1+ tan^2 t)^(3/2) ] / [ (sec^3 t) / (4c cos3t) ] = (sec^3 t) / [ (sec^3 t) / (4c cos3t) ] = 4c cos3t So, radius of curvature = 4c cos3t Have a nice day......

Answer 9

cos (t + t) = cost.cost - sint.sint
cos 2t = cos ² t - sin ² t
cos 2t = (1 - sin ² t) - sin ² t
cos 2t = 1 - 2 sin ² t

Answer 10

cos2t = cos (t + t)
= cost*cost -sint*sint
= cos^2t - sin^2t
= (1-sin^2t) -sin^2t

= 1-2sin^2t