Question Video: Solving Trigonometric Equations Involving Special Angles | Nagwa Question Video: Solving Trigonometric Equations Involving Special Angles | Nagwa

# Question Video: Solving Trigonometric Equations Involving Special Angles Mathematics • First Year of Secondary School

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Find the set of possible solutions of sin² 𝜃 − cos² 𝜃 = 0 given 𝜃 ∈ [0°, 360°).

04:19

### Video Transcript

Find the set of possible solutions of sin squared 𝜃 minus cos squared 𝜃 equals zero given that 𝜃 is greater than or equal to zero degrees but less than 360 degrees.

Now to solve this problem, what we’re gonna use is one of our trig identities. And the identity we’re gonna use is cos squared 𝜃 equals one minus sin squared 𝜃. And we’re gonna do that because we want to be left with either just sine or just cosine in our equation. So when we substitute this in, what we’re gonna get is sin squared 𝜃 minus one minus sin squared 𝜃 is equal to zero. So then if we simplify, what we’re gonna get is two sin squared 𝜃 minus one equals zero.

And it’s just worth noting here is a common mistake because what we need to watch out for is the negative in front of the parentheses because what this means is we’re gonna have negative one and then we’ve got negative one multiplied by negative sin squared 𝜃, which is gonna give us positive sin squared 𝜃. So that’s what we have: two sin squared 𝜃 minus one equals zero. So now if we add one to each side of the equation, we get two sin squared 𝜃 equals to one. And then if we divide through by two, what we’re gonna get is sin squared 𝜃 is equal to a half.

So now to solve for sin 𝜃, what we’re gonna do is take the square root of both sides of the equation. So what that’s gonna give us is that sin 𝜃 is equal to positive or negative root a half. Well, if we have the square root of one over two or the square root of a half, this is the same as the square root of one over the square root of two. Well, if we have this, then we could see that we have a surd as the denominator. And we never want a surd as a denominator, so what we do is we rationalize the denominator by multiplying by root two over root two, which is gonna give us root two over two.

Okay great, but why do we want this? Why do we do this in the first place? Well, now, what we’ve been left with is, in fact, one of our exact trig values. And the exact trig value that we know relates to this is that sin 45 is equal to root two over two. Well, great, if we use this, then what we’ve got are two values for our 𝜃, 45 degrees and negative 45. And that’s cause we had sin 𝜃 equals positive or negative root two over two. But what we can do straightaway is rule out negative 45 degrees. And that’s because it doesn’t fall in the interval we’re looking at because we’re looking in the interval where 𝜃 is greater than or equal to zero but less than 360.

Well, now what we want do is find out what the other possible values of 𝜃 are. And to do this, we’re gonna use a CAST diagram. And a CAST diagram is broken up into four quadrants A, S, T, and C. And what the A, S, T, and C stand for are where our trigonometric ratios are positive. So for A, all of the ratios in this quadrant are positive; S, only the sine ratios are positive; T, only the tangent ratios; and C, only the cosine ratios. Well, the first value we can add to our CAST diagram is 45 degrees. And that’s cause this was the one with the principal values we found.

Well then, the next value is gonna be 135 degrees. And we get that because what we can see is there actually is symmetry within our CAST diagram and we’ve got 45 degrees from the horizontal here. And we also know that it’s gonna be in this quadrant because when we looked at our sin 𝜃 value, it was either positive or negative. And in this quadrant, it’d be positive values of sin 𝜃. And because we are looking at both positive and negative values of sin 𝜃, it means that we can look at values for each of our quadrants as long as the corresponding 𝜃-values fall within our interval, which means the next value is going to be 225 degrees, which this time is 45 degrees greater than 180. So we’ve got that 45 degrees angle with the horizontal.

So then the final value in the interval that we’re looking at is 315 degrees. And again, this is 45 degrees from the horizontal. So therefore, we can say that the set of possible solutions of sin squared 𝜃 minus cos squared 𝜃 equals zero given that 𝜃 is greater than or equal to zero degrees but less than 360 degrees is 45 degrees, 135 degrees, 225 degrees, and 315 degrees.

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