Question Video: Solving Trigonometric Equations Involving Special Angles and Periodic Identities | Nagwa Question Video: Solving Trigonometric Equations Involving Special Angles and Periodic Identities | Nagwa

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

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Find the set of values of 𝜃 that satisfies cot 𝜃/sin 𝜃 = csc 𝜃, where 𝜃 ∈ [0°, 360°].

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Video Transcript

Find the set of values of 𝜃 that satisfies cot 𝜃 divided by sin 𝜃 is equal to csc 𝜃, where 𝜃 exists between zero and 360 degrees.

In order to solve this problem, we recall that cot 𝜃 is equal to cos 𝜃 over sin 𝜃 and that csc 𝜃 is equal to one over sin 𝜃. Replacing csc 𝜃 with one over sin 𝜃, we have cot 𝜃 over sin 𝜃 is equal to one over sin 𝜃. As the denominators in this equation are equal, their numerators have to be equal. cot 𝜃 must be equal to one. This means that cos 𝜃 over sin 𝜃 must be equal to one. Multiplying both sides of this equation by sin 𝜃 gives us cos 𝜃 is equal to sin 𝜃. We need to find values of 𝜃 between zero and 360 degrees where cos 𝜃 is equal to sin 𝜃.

We recall that this is true for one of our special angles, 45 degrees. We know that sin 45 degrees and cos 45 degrees are both equal to one divided by root two. This is also sometimes written as root two over two. By considering our CAST diagram, we know that sin of 𝜃 and cos of 𝜃 are both positive between zero and 90 degrees. They are both negative between 180 and 270 degrees. This means that we can calculate the second correct solution by adding 45 degrees to 180 degrees. 180 plus 45 is equal to 225.

The values of 𝜃 for which cos 𝜃 is equal to sin 𝜃 between zero and 360 degrees are 45 degrees and 225 degrees. The sin and cos of these two angles are one over root two and negative one over root two, respectively. We can, therefore, conclude that the set of values of 𝜃 that satisfies cot 𝜃 divided by sin 𝜃 is equal to csc 𝜃 are 45 degrees and 225 degrees.

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