Use the given information to determine the values of sinθ, cosθ, and tanθ.
tan2θ=7/11 ; π<2θ<3π/2
Please show work +answer. Thank you, tutor.
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.
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
I love you Reiny lol thank you sooo much. Thank you steve.
To find the values of sinθ, cosθ, and tanθ, we can use the given information that tan2θ = 7/11 and that π < 2θ < 3π/2.
First, let's determine the value of tanθ using the double angle formula:
tan2θ = (2tanθ) / (1 - tan^2θ)
Substituting the given value of tan2θ, we have:
7/11 = (2tanθ) / (1 - tan^2θ)
Next, we can solve this equation for tanθ:
7(1 - tan^2θ) = 22tanθ
7 - 7tan^2θ = 22tanθ
7tan^2θ + 22tanθ - 7 = 0
We can solve this quadratic equation for tanθ by factoring or using the quadratic formula. Factoring would be faster in this case:
(7tanθ + 1)(tanθ - 7) = 0
Setting each factor to zero, we get two potential solutions:
1) 7tanθ + 1 = 0
7tanθ = -1
tanθ = -1/7
2) tanθ - 7 = 0
tanθ = 7
Now we have two possible values for tanθ: -1/7 and 7.
Next, we can use the unit circle to determine the sign of these trigonometric functions within the range π < 2θ < 3π/2.
Since tanθ is the ratio of sinθ to cosθ, we can observe that if tanθ is positive, both sinθ and cosθ must also be positive. On the other hand, if tanθ is negative, either sinθ or cosθ (or both) must be negative.
For the range π < 2θ < 3π/2, we know that sine is negative, so sinθ < 0. Hence, we can eliminate the positive value of tanθ (tanθ = 7) because it implies sinθ and cosθ would both be positive.
Therefore, the value of tanθ within the given range is tanθ = -1/7.
To find cosθ and sinθ, we can use the trigonometric identity:
cos^2θ + sin^2θ = 1
Since we know the value of tanθ, we can use the fact that:
tan^2θ + 1 = sec^2θ
Substituting tanθ = -1/7:
(-1/7)^2 + 1 = sec^2θ
1/49 + 1 = sec^2θ
50/49 = sec^2θ
Taking the square root of both sides, we get:
sqrt(50/49) = secθ
sqrt(50) / 7 = secθ
Since secθ is the reciprocal of cosθ, we have:
cosθ = 7 / sqrt(50)
cosθ = 7sqrt(2)/10
Using the Pythagorean identity, we can now find sinθ:
sin^2θ = 1 - cos^2θ
sin^2θ = 1 - (7sqrt(2)/10)^2
sin^2θ = 1 - (98/100)
sin^2θ = 2/100
sin^2θ = 1/50
Taking the square root of both sides, we get:
sinθ = 1 / sqrt(50)
sinθ = sqrt(2)/10
To summarize, within the range π < 2θ < 3π/2, the values of sinθ, cosθ, and tanθ are:
sinθ = sqrt(2)/10
cosθ = 7sqrt(2)/10
tanθ = -1/7