Video: Solving Trigonometric Equations Involving Special Angles and Quotient Identities

Find the set of values satisfying sin πœƒ cot πœƒ = βˆ’1/2 where 0Β° ≀ πœƒ ≀ 90Β°.

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Video Transcript

Find the set of values satisfying sin πœƒ cot πœƒ equals negative one-half where πœƒ is greater than or equal to zero and less than or equal to 90 degrees.

There are two reasons we can’t immediately solve this equation. Firstly, we have the product of two trigonometric functions, sin πœƒ and cot πœƒ. Secondly, one of these trigonometric functions is a reciprocal; cot πœƒ is equal to one over tan πœƒ. So, we’ll begin by rewriting our equation as sin πœƒ times one over tan πœƒ equals negative one-half. Next, we recall that tan πœƒ is equal to sin πœƒ over cos πœƒ. Its reciprocal, one over tan πœƒ, must therefore be equal to cos πœƒ over sin πœƒ. So, we can say that sin πœƒ times cos πœƒ over sin πœƒ equals negative one-half.

We know that no matter the value of πœƒ, sin πœƒ divided by sin πœƒ will always be equal to one. And so, we need to solve the equation cos πœƒ equals negative one-half. We’ll take the inverse cos of both sides of the equation. Of course, the inverse cos of cos πœƒ is simply πœƒ, so πœƒ must be equal to the inverse cos of negative one-half. The inverse cos of negative one-half is 120 degrees. So, this is one value of πœƒ that satisfies our equation.

We do have a bit of a problem though. We’re looking for the set of values that satisfy our equation where πœƒ is greater than or equal to zero and less than or equal to 90 degrees. 120 degrees clearly doesn’t satisfy this criteria, so we’ll use the fact that cos of πœƒ is periodic to see if we can find any other solutions inside this interval. We could use a cast diagram. Alternatively, we can use the graph of 𝑦 equals cos πœƒ. It looks a little something like this.

We’re looking for values of πœƒ that satisfy our equation inside the interval zero to 90. I’ve represented that using this pink arrow. We solved the equation cos of πœƒ is equal to negative one-half. So, if we add the line 𝑦 equals negative one-half onto our diagram, we can see it looks a little something like this. It intersects the curve 𝑦 equals cos πœƒ at 120 degrees. We calculated that before. And another value here as shown. It doesn’t, however, intercept the curve of values of πœƒ greater than or equal to zero and less than or equal to 90. There are, therefore, no solutions at this interval. We use the symbol for the empty set, or the null set, to represent this.

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