posted by .

inverse cos (-rad2/2)=135 degrees
inverse sin (-rad2/2)=-45 degrees

I don't understand why cos is 135

  • trig -

    for the cosine = -sqrt 2/2 negative x axis
    means quadrant 2 or quadrant 3
    It could be either 180 + 45 or 180 - 45
    so either 135 or 225
    for the sine = -sqrt 2/2
    means quadrant 3 or quadrant 4
    that is 180 + 45 or 360 - 45
    so either 225 or 315

    225 satisfies both the sine and the cosine.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. 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 …
  2. Trig

    What is the exact value of cos 40 degrees cos 50 degrees minus sin 40 degrees sin 50 degrees do you know the formula cos(A+b) = cosAcosB - sinAsinB now compare it with your cos 40 degrees cos 50 degrees minus sin 40 degrees sin 50 …
  3. trig

    Give exact answers Sin(11pi/8) so this is sin(pi/4+pi/3) sin(pi/4)*cos(pi/3)+cos(pi/3)*sin(pi/4) rad2/2*1/2+1/2*rad2/2 which gives me rad2+rad2 all over 4 ...i don't know y im getting it wrong
  4. 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 …
  5. Trigonometry

    Please review and tell me if i did something wrong. Find the following functions correct to five decimal places: a. sin 22degrees 43' b. cos 44degrees 56' c. sin 49degrees 17' d. tan 11degrees 37' e. sin 79degrees 23'30' f. cot 19degrees …
  6. trigonometry (please double check this)

    Solve the following trig equations. give all the positive values of the angle between 0 degrees and 360 degrees that will satisfy each. give any approximate value to the nearest minute only. 1. sin2ƒÆ = (sqrt 3)/2 2. sin^2ƒÆ = …
  7. Math

    Trying to understand an inverse trig function. Evaluate Cos^-1 (sqr3/2) So I know it's 30 degrees, but wouldn't it also be 150 degrees because the Arc-cos is defined in the 2nd quadrant?
  8. Math

    y=arccos(sin(x)), find dy/dx and sketch it's graph(I guess I can do this on wolframalpha after I'm done solving the question). And by arcsin I mean inverse of the expression(written like cos^-1(sin(x)), but is not 1/cos(sinx) but the …
  9. math

    Determine the exact value of the following. a) sin300 degress b) tan (-135 degrees) c)csc150 degrees d) sec45 degrees e) cos 4pi/3 f) sin (-3pi/4)
  10. Trig

    Solve for t algebraically: inverse cos(t) = inverse sin(t). Where do I start?

More Similar Questions