Question Video: Solving Trigonometric Equations Involving Special Angles Mathematics

Find the value of πœƒ that satisfies csc πœƒ βˆ’ √2 = 0, where πœƒ ∈ (0, πœ‹/2 ).

03:25

Video Transcript

Find the value of πœƒ that satisfies csc of πœƒ minus root two equals zero, where πœƒ is between nought and πœ‹ by two.

First, let’s recall what csc of πœƒ means. We have that csc of πœƒ is equal to one over sin πœƒ. And now, we can substitute this back into the equation given in the question. And we get that one over sin πœƒ minus root two equals zero. Adding root two to both sides gives us β€” we get that one over sin πœƒ is equal to root two.

And now, if we multiply both sides by sin πœƒ and divide both sides by root two, then what we are left with is sin πœƒ is equal to one over root two. And from here, we can see that πœƒ is equal to sin inverse of one over root two. Typing this into our calculators, we get πœ‹ by four radians.

We can check this solution by drawing a right triangle with an angle of πœ‹ by four. Since this is an isosceles triangle, we can label two of the sides with length one, meaning the hypotenuse will have a length of root two.

And now, we can use SOHCAHTOA to help us find the value of sin of πœ‹ by four. Since we are finding the sine of an angle, we’re going to want to use SOH. And what this tells us is that sin of πœƒ is equal to the opposite over the hypotenuse. Now, the angle we’re interested in is πœ‹ by four. So let’s call that πœƒ.

And now, we can label on the opposite adjacent and hypotenuse with respect to this angle πœƒ. So the adjacent is the side next to the angle. The opposite is the side across from it. And the hypotenuse is the side opposite the right angle. Now, we have that sin of πœ‹ by four is equal to the opposite over the hypotenuse. So that’s one over root two. And this agrees with our answer here.

Finally, we just need to check that our value of πœƒ lies within the range between nought and πœ‹ by two. So in order to do this, let’s draw a sine graph. Here is our graph with the line 𝑦 equals sin πœƒ. Now, let’s mark on our line at πœƒ equals zero and at πœƒ equals πœ‹ by two. And we can shade the region which we’re excluding since we are only interested in values between nought and πœ‹ by two.

Now, we have that one over root two is roughly equal to 0.7. And we can use this to help us mark on the line 𝑦 equals one over root two. And so we can see that our solution for πœƒ is the point where the lines 𝑦 equals one over root two and 𝑦 equals sin πœƒ intersect. And we can see that this is at the point πœ‹ by four, which agrees with what we found earlier.

Therefore, we can be sure that our solution here is πœƒ equals πœ‹ by four.

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