Question Video: Using the Trigonometric Values of Special Angles to Find the Measure of an Angle Mathematics

If the angle πœƒ is in the standard position, cos πœƒ= βˆ’(√(2)/2) and sin πœƒ= βˆ’(√(2)/2), is it possible for πœƒ to measure 135Β°?

04:40

Video Transcript

If the angle πœƒ is in the standard position, cos of πœƒ equals negative root two over two, and sin of πœƒ equals negative root two over two, is it possible for πœƒ to measure 135 degrees?

Let’s consider the unit circle. Remember, the length of the line joining the origin to any point on this circle is one unit. It’s a circle with a radius of one. The circle intersects the axes at one, zero; zero, one; negative one, zero; and zero, negative one.

Remember since we’re measuring in an anticlockwise direction, we can start at the point one, zero at zero degrees. The second point corresponds to 90 degrees. The third 180 degrees. The fourth 270. And when we get back to the start, we’re at the angle 360 degrees.

Now, let’s choose any point on the circumference of this circle. We’ll call this ordered pair π‘Ž, 𝑏. We can then create a right-angle triangle from this point. This represents the angle πœƒ. And this will help us decide if sin πœƒ and cos πœƒ for this angle are positive or negative.

Now, we already know that, in this right-angle triangle, the hypotenuse, which is the radius, has a value of one. The side opposite to the angle πœƒ is 𝑏-units since we chose the 𝑦-coordinate to be 𝑏. The adjacent side has a value of π‘Ž-units since we chose the π‘₯-coordinate to be π‘Ž.

Now, let’s refer back to our definition for sine and cosine. Sin of πœƒ is equal to opposite divided by hypotenuse. And cos of πœƒ is equal to adjacent divided by hypotenuse. For this triangle then, sin of πœƒ is equal to 𝑏 over one since the length of the opposite side in this triangle is 𝑏 and the length of the hypotenuse is one. This simplifies simply to 𝑏. Similarly, cos of πœƒ is going to be equal to π‘Ž over one since we said that the length of the adjacent side is π‘Ž. π‘Ž over one simplifies simply to π‘Ž.

Since both π‘Ž and 𝑏 lie in the first quadrant, we know they’re both positive values. We therefore know that any value of πœƒ between zero and 90 degrees β€” in other words, any value of πœƒ that lies in this first quadrant, both sin of πœƒ and cos of πœƒ must be greater than zero. They’re positive.

Let’s look back to the original question. Both our values of sin πœƒ and cos πœƒ are negative. Therefore, we need to find the quadrant that πœƒ lies in for which this is the case. Let’s try the second quadrant. In this quadrant, we can choose an ordered pair and call it negative π‘Ž, 𝑏. Once again, we’ll use our definitions for sine and cosine to work out if the values of sine and cosine are positive or negative.

The length of the opposite side in this triangle is 𝑏. So once again, sin of πœƒ is equal to 𝑏 over one, which simplifies to 𝑏. Cos of πœƒ however is equal to negative π‘Ž over one, which simplifies to negative π‘Ž. This is because the adjacent side of this triangle can be calculated using the π‘₯-coordinate from this ordered pair. This means then that, in the second quadrant, sin of πœƒ is greater than zero. It’s positive. And cos of πœƒ is less than zero. It’s negative.

In the third quadrant, our ordered pair is given by negative π‘Ž, negative 𝑏. In this case, the value of sin πœƒ can be found by dividing negative 𝑏 by one, which is negative 𝑏. Cos πœƒ is found by calculating negative π‘Ž divided by one, which is negative π‘Ž. In the third quadrant then, both sine and cosine are negative.

Earlier, we labeled the angles in an anticlockwise direction. And we said that, in the third quadrant, the angle πœƒ is between 180 and 270 degrees. When cos of πœƒ is negative root two over two and sin of πœƒ is negative root two over two, the value of πœƒ must lie somewhere between 180 and 270 degrees.

So no, πœƒ cannot be 135 degrees.

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