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œƒ, given πœƒ is in standard position and its terminal side passes through the point (βˆ’3/5, βˆ’4/5).

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Video Transcript

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

Since this is an π‘₯-𝑦 point, let’s think about where this will be located. In quadrant number one, π‘₯ and 𝑦 would be positive. In quadrant two, π‘₯ would be negative and 𝑦 would be positive. And in quadrant three that’s in the bottom left-hand corner, π‘₯ would be negative and 𝑦 would be negative. And in the fourth quadrant, π‘₯ would be positive and 𝑦 would be negative.

So since π‘₯ and 𝑦 are both negative, we will be in quadrant three, because we’re in the negative direction for π‘₯ and in the negative direction for 𝑦. When we create an angle, there’s an initial side and a terminal side. The terminal side is when it stops, so we’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œƒ in standard position in a rectangular coordinate system intersects the unit circle is cos πœƒ, and the 𝑦-coordinate is sin πœƒ. 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œƒ must be equal to negative four-fifths and cos of πœƒ must be equal to negative three-fifths.

Now tangent of πœƒ will be equal to the sin of πœƒ divided by the cos of πœƒ, 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 denominator’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œƒ is equal to four-thirds.

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