Pre calc

Given that tanθ = 2 root 10 over 9 and cscθ < 0 , find the exact value of cos(θ − pie over 4) .


I solved for the other side and got 11. After this idk what to do I was thinking using
cos(a-b)=cos(a)cos(b)-sin(a)sin(b) but idk how to witht the pie/4 someone please help

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  3. 9
asked by Samantha
  1. tanØ = 2√10/9
    recall tanØ = y/x in the corresponding right-angled triangle
    we are also given that cscØ is negative, so Ø must be in III

    then y = -2√10 and x = -9
    r^2 = 40 + 81 =121
    r = 11


    x = -9, y = -2√10, r = 11
    sinØ = -2√10/11, cosØ = -9/11

    cos(Ø - π/2)
    = cosØcos π/2 + sinØsin π/2
    = (-9/11)(0) + (-2√10/11)(1)
    = -2√10/11

    π/2 radians = 90°
    You should know the trig functions of sine, cosine, and tangents of the main angles .
    30°, 60°, 90° from the standard right-angled triangle with corresponding sides 1 : √3 : 2
    and the 45 - 45 - 90 triangle with sides
    1 : 1 : √2
    trig functions of 0, 90, 180 270 and 360 you should know by looking at their curves.

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    posted by Reiny
  2. well, the sides could be 1, (2/9)sqrt10 and hypotenuse 11/9

    or
    9, 2 sqrt 10, 11 with 2 sqrt 10 opposite θ

    we know sin θ <0 so θ in quadrant 3 or 4
    since tan is +, must be quadrant 3

    now actually
    cos(a-b)=cosa cosb + sina sinb
    cos a = 9/11
    cos b = 1/sqrt 2
    sin a = -(2/11) sqrt10
    sin b = 1/sqrt2
    so
    9/(11 sqrt 2) - (2/11)sqrt 5
    sin a = -2 sqrt 10

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    posted by Damon
  3. but she said pi/4 :)

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    posted by Damon
  4. oh, cos also - in quadrant 3
    cos a = -9/11

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    posted by Damon

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