1. ## Solve Trigonometric Equations

So I just have a question that I am not sure about.

Why is the maximum possible value for sinx 1 and then there are no solutions?

could someone please try their best to explain this concept to me and the reason why this is true?

2. Originally Posted by skeske1234
So I just have a question that I am not sure about.

Why is the maximum possible value for sinx 1 and then there are no solutions?

could someone please try their best to explain this concept to me and the reason why this is true?

no solutions to what equation ?

3. ## Definition of sine and cosine

Hello skeske1234
Why is the maximum possible value for sinx 1
See the attached diagram.

Imagine a point $\displaystyle P$
that starts at the point $\displaystyle (1, 0)$ on the $\displaystyle x$-axis, and moves anticlockwise around a circle, centre $\displaystyle (0, 0)$, radius $\displaystyle 1$ unit.

As it does so, the line $\displaystyle OP$
rotates through an angle $\displaystyle \theta$, where $\displaystyle \theta = 0$ initially, and $\displaystyle \theta = 360^o = 2\pi$ radians when $\displaystyle OP$ has made one complete rotation.

Then the definition of $\displaystyle \sin\theta$
is the $\displaystyle y$-coordinate of $\displaystyle P$ at any moment: $\displaystyle PT$ in my diagram. (And $\displaystyle \cos\theta$ is the $\displaystyle x$-coordinate of $\displaystyle P = OT$.)

Since P is restricted to lying on the circle, its $\displaystyle x$-
and $\displaystyle y$-coordinates always lie between $\displaystyle -1$ and $\displaystyle +1$. Hence the maximum and minimum values of $\displaystyle \sin\theta$ and $\displaystyle \cos\theta$ are $\displaystyle +1$ and $\displaystyle -1$ respectively.

and then there are no solutions?
Do you mean no solutions to the equation $\displaystyle \sin x = 1$? Because if you do, that's not true. As you'll see from my diagram, $\displaystyle \sin 90^o = 1$ (and so is $\displaystyle \sin 450^o$, $\displaystyle \sin 810^o$, ... and so on.)