Question Video: Finding the Value of a Trigonometric Function of an Angle given the Coordinates of the Point of Intersection of the Terminal Side and the Unit Circle | Nagwa Question Video: Finding the Value of a Trigonometric Function of an Angle given the Coordinates of the Point of Intersection of the Terminal Side and the Unit Circle | Nagwa

# Question Video: Finding the Value of a Trigonometric Function of an Angle given the Coordinates of the Point of Intersection of the Terminal Side and the Unit Circle Mathematics • First Year of Secondary School

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Find tan π, given π is in standard position and its terminal side passes through the point (β3/5, β4/5).

02:03

### Video Transcript

Find the tangent of π, given π is in standard position and its terminal side passes through the point negative three-fifths, negative four-fifths.

Since this is an π₯-π¦ point, letβs think about where this will be located. In quadrant number one, π₯ and π¦ would be positive. In quadrant two, π₯ would be negative and π¦ would be positive. And in quadrant three thatβs in the bottom left-hand corner, π₯ would be negative and π¦ would be negative. And in the fourth quadrant, π₯ would be positive and π¦ would be negative.

So since π₯ and π¦ are both negative, we will be in quadrant three, because weβre in the negative direction for π₯ and in the negative direction for π¦. When we create an angle, thereβs an initial side and a terminal side. The terminal side is when it stops, so weβre gonna stop in quadrant three. And it says that this terminal side passes through the point negative three-fifths, negative four-fifths.

The π₯-coordinate of the point where the terminal side of an angle measuring π in standard position in a rectangular coordinate system intersects the unit circle is cos π, and the π¦-coordinate is sin π. Since the angle is in standard position and its terminal side intersects the unit circle at a point with the coordinate of negative three-fifths, negative four-fifths, sin of π must be equal to negative four-fifths and cos of π must be equal to negative three-fifths.

Now tangent of π will be equal to the sin of π divided by the cos of π, so we have negative four-fifths divided by negative three-fifths. And when we divide fractions, we actually multiply by the reciprocal, so we keep our negative four-fifths, but instead of dividing by the denominator, we multiply by the denominatorβs reciprocal, so we flip it.

And now we multiply. The fives cancel and the two negatives cancel to become a positive, so we get four-thirds. Therefore, the tangent of π is equal to four-thirds.

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