Given sin s = 3/5 and sin t = -12/13, s in quadrant I and t in quadrant III. Find sin (s+t), cos (s+t)?

Given sin s = 3/5 and sin t = -12/13, s in quadrant I and t in quadrant III. Find sin (s+t), cos (s+t), and tan (s+t).

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  • 1 decade ago
    Favorite Answer

    sin s = 3/5 : sin^2 s = 9/25: 1 - cos^2 s = 9/25: cos s = sqrt(1 - 9/25)

    cos s = 4/5 positive because cos is positive in quad I

    sin t = -12/13:sin^2 t = 144/169: 1 - cos^2 t = 144/169:

    cos t = sqrt(1 - 144/169) = sqrt(25/169)

    cos t = - 5/13 negative because cos is negative in 3rd Q

    sin(s + t) = sin s cos t + cos s sin t

    (3/5) (- 5/13) +(4/5)( -12/13)

    sin(s + t) = -63/65

    tan(s + t) = [ tan s + tan t] / [1 - tan s tan t]

    you can do the math

  • deeds
    Lv 4
    4 years ago

    For perspective u: cos u=-4/5; all of us understand that sin^2u+cos^2u=a million as a result,sin^2u=a million-(sixteen/25)=9/25; sin u=sqrt 9/25 =-3/5 ( damaging fee taken for third quadrant) For perspective v: sec^2v=a million+tan^2v; =a million+(a hundred and forty four/25) =169/25 sec v=sqrt 169/25=-13/5 (damaging in third quadrant) So cos v=-5/13 Now use those values in the formulation: sin(u+v)=siu cosv+cosu sinv

  • Como
    Lv 7
    1 decade ago

    sin (s + t)

    sin s cos t + cos s sin t

    (3/5) (-5/13) + (4/5) (-12/13)

    - 63 / 65

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