i need solutions please

[nessmell]

in ]-π , 2π[

 

[Deleted member 4993]

View attachment 21839
in ]-π , 2π[

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520​

Please share your work/thoughts about this problem.

 

[nessmell]

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520​

Please share your work/thoughts about this problem.

i m sorry i m new here ... i don't know where to start

 

[Deleted member 4993]

i m sorry i m new here ... i don't know were to start

Have you studied "unit circle"?

 

[HallsofIvy]

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.

 

[pka]

View attachment 21839
in ]-π , 2π[

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 .

 

[nessmell]

Have you studied "unit circle"?

yes

 

[nessmell]

Because of some confusion I looked HERE .

thank you

 

[HallsofIvy]

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]

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

 

[Deleted member 4993]

yes

In which quadrant (I or II or III or IV), the value of "cosine" is negative (< 0)?

 

[nessmell]

In which quadrant (I or II or III or IV), the value of "cosine" is negative (< 0)?

 

[HallsofIvy]

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)