How do you use the half-angle identity to find the exact value of cos(17π/12)?

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How do you use the half-angle identity to find the exact value of cos(17π/12)?

original question: "Use a half-angle formula to find the exact value of cos 17π/12"

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Answer 1 · 2017-04-02T18:33:28

As demonstrated below

Explanation:

I derive formulae as I typically do and I hope you find it helpful in spite of being a little long.

First, we find the half-angle formula for the cosine. We know that
$cos (2 x) = {cos}^{2} (x) - {sin}^{2} (x) = {cos}^{2} (x) - (1 - {cos}^{2} (x)) = 2 {cos}^{2} (x) - 1$ $cos(2x) = cos^2(x) - sin^2(x) = cos^2(x) - (1 - cos^2(x)) = 2cos^2(x) - 1$ so solving for $cos (x)$ $cos(x)$ we get
$cos (x) = \pm \sqrt{\frac{cos (2 x) + 1}{2}}$ $cos(x) = +- sqrt((cos(2x) + 1)/2)$ ,
where we have to determine the sign later. Replacing $x$ $x$ with $\frac{x}{2}$ $x/2$ we get a half-angle formula
$cos (\frac{x}{2}) = \pm \sqrt{\frac{cos (x) + 1}{2}}$ $cos(x/2) = +- sqrt((cos(x) + 1)/2)$ .

If we let $x = \frac{17 π}{6}$ $x = (17 pi)/6$ , then we get that
$cos (\frac{17 π}{12}) = \pm \sqrt{\frac{cos (\frac{17 π}{6}) + 1}{2}}$ $cos((17 pi) /12) = +- sqrt((cos((17 pi)/6) + 1)/2)$ .

Let us compute $cos (\frac{17 π}{6})$ $cos((17 pi)/6)$ by observing that $\frac{17 π}{6} = \frac{18 π}{6} - \frac{π}{6} = 3 π - \frac{π}{6}$ $(17 pi)/6 = (18 pi)/6 - pi/6 = 3pi - pi/6$ . Rewriting fractions of $π$ $pi$ to a sum of an integer multiple of $π$ $pi$ and a smaller fraction of $π$ $pi$ is often useful for finding exact values in trigonometric tables.

Using that
$cos (2 π + y) = cos (y)$ $cos(2pi + y) = cos(y)$ ,
$cos (π + y) = - cos (y)$ $cos(pi + y) = -cos(y)$ ,
and
$cos (- y) = cos (y)$ $cos(-y) = cos(y)$
we find that
$cos (\frac{17 π}{6}) = cos (\frac{18 π}{6} - \frac{π}{6}) = cos (2 π + π - \frac{π}{6}) = - cos (- \frac{π}{6}) = - cos (\frac{π}{6}),$ $cos((17pi)/6) = cos((18pi)/6 - pi/6) = cos(2pi + pi - pi/6) = -cos(-pi/6) = -cos(pi/6),$
which we can find in standard tables (or the unit circle) to be equal to $- \frac{\sqrt{3}}{2}$ $-sqrt(3)/2$ . Plugging into our half-angle formula we get that
$cos (\frac{17 π}{12}) = \pm \frac{\sqrt{- \frac{\sqrt{3}}{2} + 1}}{2}$ $cos((17 pi) /12) = +- sqrt(-sqrt(3)/2 + 1)/2$ .

What remains now is to figure out the sign of $cos (\frac{17 π}{12})$ $cos((17 pi)/12)$ . We find that the angle is in third quadrant by observing that
$π < \frac{17 π}{12} < \frac{18 π}{12} = \frac{3 π}{2}$ $pi<(17 pi)/12<(18 pi)/12=(3 pi)/2$ .
There, the cosine is negative (look up in a table or book, or look at the unit circle).

Therefore, we must choose the negative sign, and conclude that
$cos (\frac{17 π}{12}) = - \frac{1}{2} \sqrt{- \frac{\sqrt{3}}{2} + 1}$ $cos((17 pi) /12) = - 1/2sqrt(-sqrt(3)/2 + 1)$ .

Answer 2 · 2017-04-04T01:05:55

$cos (\frac{17 π}{12}) = - \frac{\sqrt{2 - \sqrt{3}}}{2}$ $cos ((17pi)/12) = - sqrt(2 - sqrt3)/2$

Explanation:

Use trig table and unit circle:
$cos (\frac{17 π}{12}) = cos (\frac{- 7 π}{12} + 2 π) = cos (\frac{- 7 π}{12}) = cos (\frac{7 π}{12})$ $cos ((17pi)/12) = cos ((-7pi)/12 + 2pi) = cos ((-7pi)/12) = cos ((7pi)/12)$
$= cos (\frac{π}{12} + \frac{π}{2}) = - sin (\frac{π}{2})$ $= cos (pi/12 + pi/2) = - sin (pi/2)$
Find sin (pi/12) by applying trig identity -->
$2 {sin}^{2} a = 1 - cos 2 a$ $2sin^2 a = 1 - cos 2a$
In this case:
$2 {sin}^{2} (\frac{π}{12}) = 1 - cos (\frac{π}{6}) = 1 - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2}$ $2sin^2 (pi/12) = 1 - cos (pi/6) = 1 - sqrt3/2 = (2 - sqrt3)/2$
${sin}^{2} (\frac{π}{12}) = \frac{2 - \sqrt{3}}{4}$ $sin^2 (pi/12) = (2 - sqrt3)/4$
$sin (\frac{π}{12}) = \pm \frac{\sqrt{2 - \sqrt{3}}}{2}$ $sin (pi/12) = +- sqrt(2 - sqrt3)/2$
Since $sin (\frac{π}{12})$ $sin (pi/12)$ is positive, take the positive value.
Finally,
$cos (\frac{17 π}{12}) = - sin (\frac{π}{12}) = - \frac{\sqrt{2 - \sqrt{3}}}{2}$ $cos ((17pi)/12) = - sin (pi/12) = - sqrt(2 - sqrt3)/2$

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How do you use the half-angle identity to find the exact value of cos(17π/12)?

original question: "Use a half-angle formula to find the exact value of cos 17π/12"

2 Answers

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