i need solutions please

nessmell

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in ]-π , 2π[
 

Subhotosh Khan

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nessmell

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Subhotosh Khan

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HallsofIvy

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pka

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What definition of "cosine" are you using? The standard "trigonometry" definition, "near side over hypotenuse", doesn't apply here because that is never negative. So you must have been given another definition in order to be able to do this problem, perhaps the "circular function" definition, https://www.mathplanet.com/education/algebra-2/trigonometry/circular-functions.
Because of some confusion I looked HERE .
 

HallsofIvy

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Can you do the problem now? I presume you know that an "equilateral triangle" has a all sides the same length (let's call th\fat "s") and all angles the same measure (so 180/3= 60 degrees). The line from one vertex perpendicular to the opposite side bisects that side and bisects the angle- so it divides the equilateral triangle into two 30- 60- 90 right triangles. The hypotenuse of each of those right triangles is s while the side opposite the 30 degree angle is s/2. By the Pythagorean theorem, the side opposite the 60 degree triangle has length t that satisfies \(\displaystyle t^2+ (side/2)^2= s^2\) so \(\displaystyle t^2= s^2- (s/2)^2= s^2- s^2/4= s^2(1- 1/4)= s^2(3/4)\) so that \(\displaystyle t= \frac{\sqrt{3}}{2}s\). The side opposite the 60 degree angle (so next to the 30 degree angle) divided by the hypotenuse is \(\displaystyle \frac{\sqrt{3}}{2}\).
 

nessmell

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Can you do the problem now? I presume you know that an "equilateral triangle" has a all sides the same length (let's call th\fat "s") and all angles the same measure (so 180/3= 60 degrees). The line from one vertex perpendicular to the opposite side bisects that side and bisects the angle- so it divides the equilateral triangle into two 30- 60- 90 right triangles. The hypotenuse of each of those right triangles is s while the side opposite the 30 degree angle is s/2. By the Pythagorean theorem, the side opposite the 60 degree triangle has length t that satisfies \(\displaystyle t^2+ (side/2)^2= s^2\) so \(\displaystyle t^2= s^2- (s/2)^2= s^2- s^2/4= s^2(1- 1/4)= s^2(3/4)\) so that \(\displaystyle t= \frac{\sqrt{3}}{2}s\). The side opposite the 60 degree angle (so next to the 30 degree angle) divided by the hypotenuse is \(\displaystyle \frac{\sqrt{3}}{2}\).
thank you ... we have -√3/2 not √3/2
 

HallsofIvy

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Cute! So now you know, from post, from post #9. that cos(30)=√32cos(30)=32. Now you know that cosine is negative in the second and third quadrants. cos(180- 30)= -cos(30)
 
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