# Video: Solving Trigonometric Equation Involving Special Angles

Tim Burnham

Find the set of values satisfying sin 𝑥 = −√(2)/2, where 0 ≤ 𝑥< 2𝜋.

04:54

### Video Transcript

Find the set of values satisfying sin 𝑥 equals negative root two over two, where zero is less than or equal to 𝑥 is less than two 𝜋.

Now the range that we’ve been given in this format, zero is less than or equal to 𝑥 is less than two 𝜋, is telling us that we need to give our answers in radians rather than degrees. And also we often use our calculator to work out the sine of angles. But there are certain angles that we should remember the sine or cosine or tangent of just from memory.

Let’s consider this right triangle with two sides equal in length of one unit, and we’ll call this angle 𝑥. We can use the Pythagorean theorem to work out the length of the hypotenuse; it’s the square root of one squared plus one squared, so that’s the square root of two.

Now if we want to work out the sine of that angle, this side here is the opposite side, this is the adjacent side, and this is the hypotenuse. Now sine of an angle is the opposite divided by the hypotenuse.

So that’s one over root two, but I’ve got an irrational number in the denominator. So if I multiply by root two over root two, which after all is just one — I’m not changing the magnitude of that number; I’m just changing the format that we present it in — I’ll end up with one times root two on the top, which is root two, and root two times root two on the bottom, which is just two. So sine of this particular angle here is root two over two.

Now the way we set this triangle up, we’ve got an isosceles triangle. So those two sides of both length one, the angle 𝑥 is gonna be 45 degrees or 𝜋 by four radians. Well that’s all very interesting, but that’s not the angle we’re looking for. We’re looking for an angle that has got a sine of negative root two over two, not positive root two over two.

Let’s consider angles on the unit circle. The range we were given for our answers was zero to two 𝜋, so that’s basically one revolution within the unit circle.

Now if we map out the angle 𝜋 by four on our unit circle, which was the angle that we said corresponded to a sine of root two over two, then we generate this little triangle here which has a hypotenuse of length one because it’s the unit circle, and the height of that triangle is the sine of angle 𝑥.

Now remember 𝑥 is 𝜋 by four radians, so the sine of 𝜋 by four radians is root two over two. Now if I reflect that triangle in the 𝑦-axis, again it’s gonna have a hypotenuse of one and it’s gonna have a height of root two over two. But the angle that that corresponds to isn’t just up to 𝑥; it’s all the way round to here. And that is three 𝜋 by four.

So we’re saying sine of three 𝜋 by four is equal to root two over two. Well this is all very interesting, but we’ve spectacularly failed to answer our question because we’re looking for angles which have a sine of negative root two over two, not root two over two.

But look if I take that second triangle and I reflect that in the 𝑥-axis, because the height is now below the 𝑥-axis, it’s gone negative. So that height is negative root two over two.

So let’s see what angle that corresponds to. It’s this one here that goes beyond the original 𝑥, beyond the one we were just looking at, and all the way down to here. And that is gonna be an angle of five 𝜋 by four.

It’s basically 𝜋, which is four 𝜋 by four, plus an extra 𝜋 by four. So that’s the first of our solutions. Now if we reflect that in the 𝑦-axis again, that reflected triangle will also have a height of negative root two over two because it’s below the 𝑥-axis. And this corresponds to an angle that goes all the way round to here, and that’s seven 𝜋 by four.

And that’s our second answer. So in one revolution from 𝑥 between or equal to zero and up to but not quite including two 𝜋, we found two solutions which give us a sine value of negative root two over two.

And putting that into set notation as the question asked for, we’ve got two solutions. And the set of solutions is, the first one is five 𝜋 by four and the second one is seven 𝜋 by four.

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