Video: Using the Tangent Sum or Difference of Angles Identity to Solve a Trigonometric Equation

Find the solution set of π‘₯, given (tan(π‘₯) βˆ’ tan(64Β°))/(1 + tan(π‘₯) tan(64Β°)) = 1 where 0Β° ≀ π‘₯ ≀ 360Β°.

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Video Transcript

Find the solution set of π‘₯ given tan of π‘₯ minus tan of 64 [degrees] over one plus tan of π‘₯ times tan of 64 [degrees] equals one, where π‘₯ is between zero and 360 degrees.

First, let’s recall the tan compound angle formula for tan of 𝐴 minus 𝐡. We have that tan of 𝐴 minus 𝐡 is equal to tan of 𝐴 minus tan of 𝐡 over one plus tan of 𝐴 times tan of 𝐡. Now, if we compare the right-hand side of this formula with the left-hand side of the equation given in the question, we can see that they look incredibly similar. Now, we can compare these two. And if we let 𝐴 equal π‘₯ and 𝐡 equals 64, then we’ll see that they are actually the same.

So let’s substitute 𝐴 equals π‘₯ and 𝐡 equals 64 into this equation. This gives us that tan of π‘₯ minus 64 is equal to tan of π‘₯ minus tan of 64 over one plus tan of π‘₯ times tan of 64. And the right-hand side of this equation is identical to the left-hand side of the equation given in the question. And so that means this is also equal to one. And so from this, we get the equation tan of π‘₯ minus 64 equals one.

Now, when solving this, we need to find values of π‘₯, which are between nought and 360 degrees. However, in our equation, we have tan of π‘₯ minus 64. So in order to find the range of values for π‘₯ minus 64, we can subtract 64 from each part of this inequality. So this gives us that π‘₯ minus 64 is between minus 64 and 296 degrees.

Now, we can make this look a little bit nicer if we let πœƒ equal π‘₯ minus 64. And so therefore, the range of values for πœƒ is minus 64 to 296 degrees. And we can also substitute πœƒ into our equation with tan. And so now, tan of πœƒ is equal to one is the equation we’re trying to solve. And we need πœƒ in the range between minus 64 and 296 degrees. Rearranging this equation, we get that πœƒ is equal to tan inverse of one and this is also equal to 45 degrees.

Now, we can check this using a right-angled triangle. We need a right-angled triangle with an angle of 45 degrees. So we get a triangle with two sides of length one and the hypotenuse of length root two.

And now, we can use SOHCAHTOA. Since we’re trying to find tan of an angle, we’re going to use TOA. What this means is that tan of πœƒ is equal to the opposite over the adjacent. It’s the angle we’re interested in, which we’ll call πœƒ is 45 degrees. And now, we can label on the opposite, adjacent, and hypotenuse of the triangle. And from this, we can see that tan of 45 degrees is equal to one over one. And then, obviously, this is equal to one. So this agrees with what we got. So our answer πœƒ equals 45 degrees is correct.

Now, we remember that πœƒ is in the range of minus 64 to 296 degrees. So we need to find all solutions which are in this range. In order to do this, we can draw a tan graph. Here is our tan graph. Now, we can label on tan of πœƒ, which is also equal to one. And any point where this coincides with our tan of πœƒ line will give us a solution for πœƒ.

The first point where it intercepts with the tan πœƒ line is the angle we already found of 45 degrees. Now, we can see the second point, where it intercepts between 180 degrees and 270 degrees. In order to find what this angle is, we can use the fact that the graph repeats itself every 180 degrees. And so that means that the distance from the origin to 45 degrees is gonna be the same as the distance from 180 degrees to the angle we’re trying to find.

Now the distance from the origin to 45 degrees is obviously 45. So therefore, the distance from 180 to the angle we’re trying to find is also 45 degrees. And so the angle we’re trying to find is 180 plus 45, which also equals 225 degrees. So now, we have two solutions for πœƒ. And since our dashed horizontal line at one does not intercept the graph anywhere else between minus 64 and 296 degrees, we have found all the solutions within our range.

So now, we can use our two solutions of 45 degrees and 225 degrees to find the solution set of π‘₯. Now, what we need to do is substitute the values of 45 and 225 into our equation which we have for πœƒ and π‘₯. So that’s πœƒ equals π‘₯ minus 64.

Let’s start with 45 degrees. So we have 45 is equal to π‘₯ minus 64. Now, we add 64 to both sides in order to make π‘₯ the subject. And we get π‘₯ equals 109 degrees. Now for 225 degrees, we have that 225 is equal to π‘₯ minus 64. And now, we add 64 to both sides. And this gives us a solution of 289 degrees. And so these are our two solutions. So we obtain the solution set of 109 degrees and 289 degrees.

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