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Video: Finding the General Solution of the Sine Function Involving Special Angles

Alex Cutbill

What is the general solution of sin 𝜃 = √2/2?

04:03

Video Transcript

What is the general solution of sin 𝜃 equals the square root of two over two?

Well, one solution is 𝜋 by four radians. You might know this because 𝜋 by four is a special angle, and so you might know that the value of sin 𝜋 by four is root two over two. If you didn’t, then you could use your calculator and put in arcsin or sin superscript minus one of root two over two, and it would give you 𝜋 by four radians, assuming that you had it in the correct radian mode or often degrees or something else.

But of course this is only one solution to the equation and we’d like the general solution, so how do we find that?

Take a look at the graph of 𝑦 equals sin 𝜃. If we draw the line 𝑦 equals root two over two, we see it intersects the graph of sin infinitely often. As expected, one of these intersections occurs when 𝑥 is 𝜋 by four, which is one of the solutions we found.

We can use this solution and the symmetries of the graph of sin to find all the other solutions. For example, can we see which value of 𝜃 corresponds to this intersection?

We can see that this value of 𝜃 is as far away from 𝜋 as 𝜋 by four is away from zero, so this solution is 𝜋 minus 𝜋 by four, which we could choose to write as three 𝜋 by four.

And if you’re not convinced just by looking at the graph that this is true, we can use the fact that sin 𝜋 minus 𝜃 is equal to sin 𝜃. So sin 𝜋 minus 𝜋 by four is sin 𝜋 by four, just replacing 𝜃 by 𝜋 by four. Because sin 𝜋 by four is root two over two as we showed earlier, sin three 𝜋 by four is root two over two as well.

So now we have two solutions to our equation, but we can still see on the graph there are many more. However, while these two solutions aren’t the only solutions, they are the only solutions between zero and two 𝜋.

The other solutions all arise because of the periodicity of the function sin. For example, this solution is just two 𝜋 more than the solution we had before, which was 𝜋 by four, and of course there’ll be another solution in another two 𝜋, although it’s not drawn on the graph.

And the same is true if we subtract two 𝜋. In fact, for any integer 𝑛, 𝜋 by four plus two 𝜋𝑛 is a solution. The same is true for the solution three 𝜋 by four for any integer 𝑛, three 𝜋 over four plus two 𝜋𝑛 is a solution to our equation, and you should be able to see from the graph that any solution to this equation has one of those two forms.

So let’s write these two forms together, and that will be our general solution. Combining the two expressions we had before, we get the general solution that 𝜃 is 𝜋 by four plus two 𝜋𝑛 or three 𝜋 by four plus two 𝜋𝑛, where 𝑛 is an integer.

Alternatively, we could choose to leave 𝜋 minus 𝜋 by four unsimplified. This makes it easier to see the relationship it has with sin inverse of root two over two, which is 𝜋 by four, and so the same answer written in a slightly different way and simplified would be 𝜋 by four plus two 𝜋𝑛 or 𝜋 minus 𝜋 by four plus two 𝜋𝑛, where 𝑛 is an integer.