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kakasveta [241]
3 years ago
13

What is the exact value of tan ( 5 pi/8 )

Mathematics
1 answer:
Naddika [18.5K]3 years ago
4 0
<h3>Answer: Choice D) \tan\left(\frac{5\pi}{8}\right) = - \sqrt{\frac{2+\sqrt{2}}{1-\sqrt{2}}}\\\\\\</h3>

===================================================

Explanation:

Make sure your calculator is in radian mode. Use your calculator to find that tan(5pi/8) = -2.41421 which is approximate.

Since the value is negative, this means the answer is between choices C and D. You can use your calculator to compute those expressions given and you should find it matches with choice D.

----------------------

Further explanation:

We can apply the half angle identity for tangent like so

\tan\left(\frac{x}{2}\right) = \pm \sqrt{\frac{1-\cos(x)}{1+\cos(x)}}\\\\\\\tan\left(\frac{5\pi/4}{2}\right) = -\sqrt{\frac{1-\cos(5\pi/4)}{1+\cos(5\pi/4)}}\\\\\\\tan\left(\frac{5\pi}{8}\right) = -\sqrt{\frac{1-\left(-\frac{\sqrt{2}}{2}\right)}{1+\left(-\frac{\sqrt{2}}{2}\right)}}\\\\\\

Simplifying further, we get

\tan\left(\frac{5\pi}{8}\right) = -\sqrt{\frac{1+\frac{\sqrt{2}}{2}}{1-\frac{\sqrt{2}}{2}}}\\\\\\\tan\left(\frac{5\pi}{8}\right) = -\sqrt{\frac{2+\sqrt{2}}{1-\sqrt{2}}}\\\\\\

In the last step, I multiplied top and bottom of the outer fraction by 2 to clear out the denominators of the inner fraction.

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Use the Parabola tool to graph the quadratic function. f(x)=2x^2−12x+19 Graph the parabola by first plotting its vertex and then
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To find the vertex of our parabola, we are going to use the vertex formula: For a quadratic of the form f(x)=x^2+bx+c it vertex (h,k) is given by h= \frac{-b}{2a}, k=f(h).
We can infet from our parabola that a=2 and b=-12. So lets replace those values in our formula:
h= \frac{-b}{2a}
h= \frac{-(-12)}{2(2)}
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The vertex (h,k) of our parabola is (3,1)

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3 years ago
Need help finding n forgot how to do this
andreyandreev [35.5K]

Answer:

n  =                                    3\sqrt{2}

There are two best ways to solve this.

using cosine method:

cos(n) =  \frac{adjacent}{hypotenuse}

cos(60) = \frac{adjacent}{6\sqrt{2} }

adjacent,n = cos(60) * 6\sqrt{2}

n  =                                    3\sqrt{2}

using sine method:

sin(n) = \frac{opposite}{hypotenuse}

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n  = sin(30) * 6\sqrt{2}

n = 3\sqrt{2}

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