Question

How do you solve `sin^2theta - cos^2theta = 0`?

Answer 1

`pi/4 + kpi`
`(3pi)/4 + kpi`

Explanation:

From the trig identity:
`cos^2 t - sin^2 t = cos 2t`, we get:
`sin^2 t - cos^2 t = - cos2t = 0`
Unit circle gives 2 solutions:
`cos 2t = 0 -->`2t = pi/2 + 2kpi`, and`2t = (3pi)/2 + 2kpi# a. #2t = pi/ + 2kpi`-->`t = pi/4 + kpi# b. #2t = (3pi)/2 + 2kpi`-->`t = (3pi)/4 + kpi#

Answer 2

Here's an alternate answer. Recall the identity `sin^2theta + cos^2theta = 1`. If you rearrange for `cos^2theta`, you should get `cos^2theta = 1-sin^2theta`.

Substituting, we have:

`sin^2theta - (1 - sin^2theta) = 0`

`sin^2theta - 1 + sin^2theta = 0`

`2sin^2theta = 1`

`sin^2theta = 1/2`

`sintheta = +- 1/sqrt(2)`

Now consider the `1-1-sqrt(2)` right triangle. This means that

`theta = pi/4 +2pin, (3pi)/4 + 2pin, (5pi)/4 + 2pin and (7pi)/4 + 2pin`

Note the period of the sine function is `2pi`.

Hopefully this helps!