Trig

posted by .

Verify:cos(360degrees-x)=cos x

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. trig

    Show that 1-cos2A/Cos^2*A = tan^2*A 1-cos2A/Cos^2*A = [Cos^2(A) - Cos(2A)]/Cos^2(A). Substitute: Cos(2A) = 2Cos^2(A) - 1: [1 - Cos^2(A)]/Cos^2(A)= Sin^2(A)/Cos^2(A) = tan^2(A)
  2. trig

    Reduce the following to the sine or cosine of one angle: (i) sin145*cos75 - cos145*sin75 (ii) cos35*cos15 - sin35*sin15 Use the formulae: sin(a+b)= sin(a) cos(b) + cos(a)sin(b) and cos(a+b)= cos(a)cos(b) - sin(a)sin)(b) (1)The quantity …
  3. trig

    it says to verify the following identity, working only on one side: cotx+tanx=cscx*secx Work the left side. cot x + tan x = cos x/sin x + sin x/cos x = (cos^2 x +sin^2x)/(sin x cos x) = 1/(sin x cos x) = 1/sin x * 1/cos x You're almost …
  4. trig

    how would you verify this trig identity (1+cos(x) / 1-cos(x)) - (1-cos(x) / 1+cos(x)) = 4cot(x)csc(x) ?
  5. Math - Solving Trig Equations

    What am I doing wrong? Equation: sin2x = 2cos2x Answers: 90 and 270 .... My Work: 2sin(x)cos(x) = 2cos(2x) sin(x) cos(x) = cos(2x) sin(x) cos(x) = 2cos^2(x) - 1 cos(x) (+/-)\sqrt{1 - cos^2(x)} = 2cos^2(x) - 1 cos^2(x)(1 - cos^2(x))
  6. TRIG!

    Posted by hayden on Monday, February 23, 2009 at 4:05pm. sin^6 x + cos^6 x=1 - (3/4)sin^2 2x work on one side only! Responses Trig please help! - Reiny, Monday, February 23, 2009 at 4:27pm LS looks like the sum of cubes sin^6 x + cos^6 …
  7. Precal

    I do not understand how to do this problem ((sin^3 A + cos^3 A)/(sin A + cos A) ) = 1 - sin A cos A note that all the trig terms are closed right after there A's example sin A cos A = sin (A) cos (A) I wrote it out like this 0 = - …
  8. trig

    find, to the nearest degree, the solution set of 4 cos^2x-1=0 in the intervak between )degrees and 360degrees.
  9. trig

    Verify the equation is an identitiy: cot(x)tan(x+pi) - sin(pi-x)cos((pi/2)-x) = cos^2x Thanks
  10. trig

    verify the identity algebraically- (5-5 cos x)(5+5 cos x)=25 (sin^x)

More Similar Questions