Trigonometric Functions

NZ Level 8 (NZC) Level 3 (NCEA) [In development]

Solve trigonometric equations with exact values

Lesson

Trigonometric equations express the situation in which we know the value of some trigonometric function but we do not know what angle leads to that value. For example, we may know that for some angle $\theta$`θ`, we have $\tan\theta=\sqrt{3}$`t``a``n``θ`=√3. The problem is to discover what values of $\theta$`θ` will make this a true statement.

Recalling the known exact values of the trigonometric functions, we would recognise that in this example $\theta=\frac{\pi}{3}$`θ`=π3 satisfies the equation. But, there are other values of $\theta$`θ` that are also solutions of the equation. This comes about because of the unit circle definition of the tangent function.

The solution $\theta=\frac{\pi}{3}$`θ`=π3 is the solution in the first quadrant. There should also be a solution in the third quadrant. It is found by adding the $\frac{\pi}{3}$π3 solution to $\pi$π so that we find the extra solution $\theta=$`θ`=$\frac{4\pi}{3}$4π3. In fact, we can keep adding or subtracting multiples of $\pi$π to the original solution in order to find as many further solutions as we require.

We say the tangent function has a *period *of $\pi$π because its values repeat at that interval.

In the cases of the sine and cosine functions, we might look for solutions to equations like $\sin\alpha=\frac{\sqrt{3}}{2}$`s``i``n``α`=√32 or $\cos\beta=\frac{1}{\sqrt{2}}$`c``o``s``β`=1√2. Again recalling the known exact values, we would recognise that $\alpha=\frac{\pi}{3}$`α`=π3 and $\beta=\frac{\pi}{4}$`β`=π4 are solutions in the first quadrant.

There should also be a second quadrant solution to $\sin\alpha=\frac{\sqrt{3}}{2}$`s``i``n``α`=√32. It is found by subtracting the first quadrant solution from $\pi$π. That is, $\alpha=\pi-\frac{\pi}{3}=\frac{2\pi}{3}$`α`=π−π3=2π3. This can be verified by reference to the unit circle diagram: it is always true that

$\sin\alpha\equiv\sin\left(\pi-\alpha\right)$`s``i``n``α`≡`s``i``n`(π−`α`)

We now have the two solutions $\alpha=\frac{\pi}{3}$`α`=π3 and $\alpha=\frac{2\pi}{3}$`α`=2π3. Any number of further solutions can be obtained by adding or subtracting multiples of $2\pi$2π to these basic solutions.

For the equation $\cos\beta=\frac{1}{\sqrt{2}}$`c``o``s``β`=1√2, there should be a fourth quadrant solution as well as the one in the first quadrant. We find it by subtracting the first quadrant $\beta$`β` from $2\pi$2π. Thus, $\beta=\frac{\pi}{4}$`β`=π4 or $\beta=\frac{7\pi}{4}$`β`=7π4 . Again, this can be verified by referring to the unit circle diagram. It is always true that

$\cos\beta\equiv\cos\left(2\pi-\beta\right)$`c``o``s``β`≡`c``o``s`(2π−`β`)

As with the sine function, as many further solutions as are needed are found by adding or subtracting multiples of $2\pi$2π to the basic solutions.

We say the sine and cosine functions are periodic with a period of $2\pi$2π because the function values recur regularly at that interval.

Find all the solutions that lie between $0$0 and $2\pi$2π of the equation $\sin2\theta=\frac{\sqrt{3}}{2}$`s``i``n`2`θ`=√32.

In the first quadrant, we have $2\theta=\frac{\pi}{3}$2`θ`=π3. It must also be true that $2\theta=\frac{2\pi}{3}$2`θ`=2π3 gives a second quadrant solution for $2\theta$2`θ`. We need to go around the circle again and find the additional solutions $2\theta=\frac{7\pi}{3}$2`θ`=7π3 and $2\theta=\frac{8\pi}{3}$2`θ`=8π3. Finally, we obtain the solutions for $\theta:$`θ`: $\theta=\frac{\pi}{6}$`θ`=π6, $\theta=\frac{\pi}{3}$`θ`=π3, $\theta=\frac{7\pi}{6}$`θ`=7π6 and $\theta=\frac{4\pi}{3}.$`θ`=4π3.

Form and use trigonometric, polynomial, and other non-linear equations

Apply trigonometric methods in solving problems