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

Question

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

Answer 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.

Answer 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

Answer 3

but she said pi/4 :)

Answer 4

oh, cos also - in quadrant 3
cos a = -9/11