Use the given information to determine the values of sinθ, cosθ​, and tanθ.

Question

tan2θ=7/11 ​; π<2θ<3π/2

Please show work +answer. Thank you, ​tutor.

Answer 1

well, you know that θ is in QIII, so

sin2θ = -7/√170
cos2θ = -11/√170

and π/2 < θ < 3π/4 so θ is in QII

Now just use the half-angle formulas to get the functions of θ.

Answer 2

tan2θ=7/11 ​; π<2θ<3π/2

so 2Ø is in quadrant III and Ø is in quadrant II

construct a right-angled triangle, with angle 2Ø, opposite side as 7, and adjacent side as 11
the hypotenuse is √170.
Remember 2Ø is in III, so
sin 2Ø= -7/√170 or -7√170/170
cos 2Ø = -11/√170 or -11√170/170

cos 2Ø = 2sin^2 Ø - 1
-11√170/170 + 1 = 2cos^2 Ø
(170 - 11√170)/170 = 2cos^2 Ø
(170 - 11√170)/340 = cos^2 Ø
cos Ø = -√(170 - 11√170)/√340 , I picked the negative, since in II the cosine is negative.

We also know that sin 2Ø = 2sinØcosØ
-7√170/170 = 2sinØ(-√(170 - 11√170)/340
2 sinØ = -7√170/170 / ( (-√(170 - 11√170)/340 )
sin Ø = (14√170/170) / (√(170 - 11√170))

Better check this algebra, I think I messed up on the sin Ø

let's do the last part in a different way: tan Ø
recall tan 2Ø = 2tanØ/(1 - tan^2 Ø)
let tanØ = x for easier typing,
7/11 = 2x/(1 - x^2)
22x = 7 - 7x^2
7x^2 + 22x - 7 = 0
x = (-22 ± √680)/14
= (-22 ± 2√170)/14
= (-11 ± √170)/7

but since Ø is in II,
x
= tan Ø
= (-11 - √170)/7

I checked on my calculator, cosØ and tanØ are correct,
check my sinØ

Answer 3

I love you Reiny lol thank you sooo much. Thank you steve.