# Video: Solving Trigonometric Equations Involving Special Angles

If 𝜃 ∈ [0°, 180°[ and sin 𝜃 + cos 𝜃 = 1, find the possible values of 𝜃.

03:40

### Video Transcript

If 𝜃 exists between zero degrees and 180 degrees, including zero but not including 180, and sin 𝜃 plus cos 𝜃 is equal to one, find the possible values of 𝜃.

If we consider the equation sin 𝜃 plus cos 𝜃 equals one, we can solve this to calculate our values of 𝜃. Our first step is to square both sides of the equation. This gives us sin 𝜃 plus cos 𝜃 all squared is equal to one squared. In order to square sin 𝜃 plus cos 𝜃, we need to multiply sin 𝜃 plus cos 𝜃 by itself. One squared is equal to one.

We can expand the double brackets or parentheses using the FOIL method. Firstly, we multiply sin 𝜃 by sin 𝜃. This gives us sin squared 𝜃. Next, we multiply sin 𝜃 by cos 𝜃. This is equal to sin 𝜃 cos 𝜃. Our next step is to multiply the inside terms, cos 𝜃 and sin 𝜃. This also gives us sin 𝜃 cos 𝜃. Finally, we multiply the last terms giving us cos squared 𝜃, as cos 𝜃 multiplied by cos 𝜃 is cos squared 𝜃.

We are now left with sin squared 𝜃 plus sin 𝜃 cos 𝜃 plus another sin 𝜃 cos 𝜃 plus cos squared 𝜃 is equal to one. This can be simplified to sin squared 𝜃 plus cos squared 𝜃 plus two sin 𝜃 cos 𝜃 is equal to one. Our next step is to use two of our trigonometrical identities. Firstly, sin squared 𝜃 plus cos squared 𝜃 is equal to one. This gives us one plus two sin 𝜃 cos 𝜃 is equal to one. We can subtract one from both sides of this equation. This gives us two sin 𝜃 cos 𝜃 is equal to zero. One of our double-angle formulae tells us that sin two 𝜃 is equal to two sin 𝜃 cos 𝜃. This means that our equation simplifies to sin two 𝜃 is equal to zero.

We are trying to solve for values between zero and 180 degrees. The graph of sin 𝜃 or sin two 𝜃 has a maximum of one and a minimum of negative one. The graph of sin two 𝜃 has a maximum value at 𝜃 equals 45 degrees and a minimum value at 𝜃 equals 135 degrees. We want the values where sin of two 𝜃 is equal to zero. This occurs at three points on our graph, at 𝜃 equal zero degrees, 90 degrees, and 180 degrees. We wanted values greater than or equal to zero degrees, but less than 180 degrees. This means that the two possible values of 𝜃 that satisfy our equation in this range are 𝜃 equals zero degrees and 90 degrees.