Solving for Double Angles

[NK8485]

I'm currently working on double angles and am running into some issues with a question on my homework.

It is asking for me to find the sinθ, cosθ, and tanθ of an angle with given information:

Tan2θ= 11/7. The angle lies in the quadrant Pi < 2θ < 3Pi/2.

I know that the angle for 2θ lies in quadrant 3 since I am given the intervals of Pi and 3pi/2. I know that Tan values are positive when the terminal side lies in quadrant 3. I also know that Tan=y/x, so a point on the terminal side of 2θ is (-7,-11). Given this I can figure out the values of sin2θ and cos2θ by using the r^2=x^2+y^2 formula which gives me sin2θ=-11/√170 and Cos2θ=-7/√170.

If I divide the intervals Pi < 2θ < 3Pi/2 by 2 I find that the angle θ lies within Pi/2 < θ < 3Pi/4, putting it in quadrant 2.

I use one of my double angle formulas to solve for the value. Cos2θ=1-2sin^2θ

So, -7/√170=1-2sin^2θ

After working through the algebra I'm left with: (7+√170)/(2√170)=sin^2θ

I know to figure out what sinθ is I need to take the square roots of both sides; however, I'm not sure how to take the square root of (7+√170)/(2√170).

After figuring out the sin value I know how to come up with the cosine and tangent values for θ, it's just the one step I'm lost on.

 

[stapel]

Thank you for showing your work and reasoning so nicely!

...find the sin, cosθ, and tanθ of an angle with given information:

\(\displaystyle \tan(\theta)\, =\, \dfrac{11}{7},\, \mbox{ with }\, \pi\, <\, 2\theta\, <\, \dfrac{3\pi}{2}\)

I know that the angle for 2θ lies in Quadrant III.... I know that tangent values are positive when the terminal side lies in Quadrant III. I also know that the tangent ratio is y/x, so a point on the terminal side of 2θ is (-7, -11). Given this I can figure out the values of sin2θ and cos2θ by using the r² = x² + y² formula which gives me:

\(\displaystyle \sin(2\theta)\, =\, -\dfrac{11}{\sqrt{\strut 170\,}}\, \mbox{ and }\, \cos(2\theta)\, =\, -\dfrac{7}{\sqrt{\strut 170\,}}\)

If I divide the intervals Pi < 2θ < 3Pi/2 by 2, I find that the angle θ lies within $\displaystyle \, \dfrac{\pi}{2}\, <\, \theta\, <\, \dfrac{3 \pi}{4},\,$ putting it in Quadrant II.

I use one of my double angle formulas to solve for the value: $\displaystyle \cos(2\theta)\, =\, 1\, -\, 2\sin^2(\theta)$

So: \(\displaystyle \,-\dfrac{7}{\sqrt{\strut 170\,}}\, =\, 1\, -\, 2\sin^2(\theta)\)

After working through the algebra I'm left with: \(\displaystyle \,\dfrac{\left(\,7\, +\, \sqrt{\strut 170\,}\,\right)}{2\, \sqrt{\strut 170\,}}\, =\, \sin^2(\theta)\)

I know to figure out what sinθ is I need to take the square roots of both sides; however, I'm not sure how to take the square root of \(\displaystyle \,\dfrac{\left(\,7\, +\, \sqrt{\strut 170\,}\,\right)}{2\, \sqrt{\strut 170\,}}.\)

I might have taken a different route, but I think I'd have ended up with similar (and equivalent) results. Sometimes, the math is just nasty.

You're wanting to take the square root of:

. . . . .\(\displaystyle \dfrac{\left(\,7\, +\, \sqrt{\strut 170\,}\,\right)}{2\, \sqrt{\strut 170\,}}\)

Probably a good first step is to rationalize the denominator:

. . . . .\(\displaystyle \left(\, \dfrac{\left(\,7\, +\, \sqrt{\strut 170\,}\,\right)}{2\, \sqrt{\strut 170\,}}\,\right)\, \left(\, \dfrac{\sqrt{\strut 170\,}}{\sqrt{\strut 170\, }}\,\right)\, =\, \dfrac{7\, \sqrt{\strut 170\,}\, +\, 170}{340}\)

Now take the square root and do some rationalising:

. . . . .\(\displaystyle \sqrt{\strut \dfrac{7\, \sqrt{\strut 170\,}\, +\, 170}{340} \,}\, =\, \dfrac{\sqrt{\strut 7\, \sqrt{\strut 170\,}\, +\, 170}}{2\, \sqrt{\strut 85\,}}\, =\, \dfrac{\sqrt{\strut 7\, \cdot\, 85\, \sqrt{2\,}\, +\, 170\, \sqrt{\strut 85\,}\,}}{2\,\cdot\,85}\)

You can do some simplifying and rearranging, but I'm not sure that there's a "nice" way of putting this.

Note: https://www.wolframalpha.com/input/?i=simplify+sqrt((7*sqrt(170)+++170)/340)