# Solving trig function (possible error in solutions)

## Homework Statement

Alright, so I am going through the solutions to one problem that has me stumped, and initial instinct tells me it is wrong, however, I figured I would get clarification.

## The Attempt at a Solution

$$sin(2t)=cos(2t)$$
$$tan(2t)=1$$
$$2t=\pi{/}4, 5\pi{/}4$$

However, the book states both positive and negative values, rather than just positive. It's either not correct, or not clicking. tan(-π/4)=-1, not 1..

I like Serena
Homework Helper
What is tan(-3pi/4)?

dynamicsolo
Homework Helper
Since your text is suggesting that there are positive and negative solutions, you are not restricted to the "principal circle" , $0 ≤ t < 2{\pi}$ . So you can "go around the circle" as many times as you like:

$$2t = \frac{\pi}{4} + k \cdot 2{\pi} , \frac{5\pi}{4} + k \cdot 2{\pi} ,$$

with k being any integer.

SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

Alright, so I am going through the solutions to one problem that has me stumped, and initial instinct tells me it is wrong, however, I figured I would get clarification.

## The Attempt at a Solution

$$sin(2t)=cos(2t)$$
$$tan(2t)=1$$
$$2t=\pi{/}4, 5\pi{/}4$$

However, the book states both positive and negative values, rather than just positive. It's either not correct, or not clicking. tan(-π/4)=-1, not 1..

Are you sure the problem isn't, $\sin^2(t)=\cos^2(t)\,?$