the unit circle
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Sorry but I don't understand could you please develop your answer?
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Have you studied "unit circle"?
Can you calculate:
\(\displaystyle cos \left(\frac{\pi}{3}\right)\) = ?......\(\displaystyle cos \left(\frac{-\pi}{3}\right)\) = ?......\(\displaystyle cos \left(\frac{5*\pi}{6}\right)\) = ?......\(\displaystyle cos \left(\frac{-5\pi}{6}\right)\) = ?......
\(\displaystyle sin \left(\frac{\pi}{3}\right)\) = ?......\(\displaystyle sin \left(\frac{-\pi}{3}\right)\) = ?......\(\displaystyle sin \left(\frac{5*\pi}{6}\right)\) = ?......\(\displaystyle sin \left(\frac{-5\pi}{6}\right)\) = ?......
I have not in school just a little bit in spare time
From the definition of the sine and cosine relative to the unit circle, we get
[MATH]0^o \le \theta \le 180^o \implies 0 \le sin(\theta) \le 1 \text { and } sin(180 - \theta) = sin(\theta).[/MATH]
[MATH]0^o \le \theta \le 90^o \implies 0 \le cos(\theta) \le 1,\ cos(\theta) = \sqrt{1 - sin^2(\theta)}.[/MATH]
[MATH]90^o < \theta \le 180^o \implies -1 \le cos( \theta ) < 0. \ cos(\theta) = - \sqrt{1 - sin^2(\theta)}.[/MATH]
In the interval [0 degrees, 180 degrees], the sine takes on the same value twice (except at 90 degrees) whereas the cosine takes on the same absolute value twice, but with opposite signs (again except at 90 degrees).
[MATH]0^o \le \theta \le 180^o \implies 0 \le sin(\theta) \le 1 \text { and } sin(180 - \theta) = sin(\theta).[/MATH]
[MATH]0^o \le \theta \le 90^o \implies 0 \le cos(\theta) \le 1,\ cos(\theta) = \sqrt{1 - sin^2(\theta)}.[/MATH]
[MATH]90^o < \theta \le 180^o \implies -1 \le cos( \theta ) < 0. \ cos(\theta) = - \sqrt{1 - sin^2(\theta)}.[/MATH]
In the interval [0 degrees, 180 degrees], the sine takes on the same value twice (except at 90 degrees) whereas the cosine takes on the same absolute value twice, but with opposite signs (again except at 90 degrees).
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Thank you so much!
Steven G
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Sin(x) is positive in quadrant 1 and 2 which covers angles from 0 to 180. So the sin(x) equaling a positive number from 0 to 180 will have two solutions.
Now cos(x) is positive in quadrant 1 and 4. So when you have to solve cos(x) equals a positive number and the angle is between 0 and 180 then there will be only one solution.
The hypotenuse is always positive. The sin(x) = opposite/hypotenuse > 0 when opposite is positive (and negative when the opposite is negative). Now the opposite is only positive when the angle is above the x axis, ie in quadrant 1 and 2.
Cos(x) = adj/hyp which is positive when the adjacent >0. This happens to the right of the y axis, ie in quadrant 1 and 4.
Now cos(x) is positive in quadrant 1 and 4. So when you have to solve cos(x) equals a positive number and the angle is between 0 and 180 then there will be only one solution.
The hypotenuse is always positive. The sin(x) = opposite/hypotenuse > 0 when opposite is positive (and negative when the opposite is negative). Now the opposite is only positive when the angle is above the x axis, ie in quadrant 1 and 2.
Cos(x) = adj/hyp which is positive when the adjacent >0. This happens to the right of the y axis, ie in quadrant 1 and 4.
Thank you I really appreciate your help!