Solve Trigonometric Equations

**skeske1234** · April 18th 2009, 07:34 AM

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?

Thanks in advance!

**skeeter** · April 18th 2009, 07:52 AM

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?

Thanks in advance!

no solutions to what equation ?

**Grandad** · April 18th 2009, 08:15 AM

Hello skeske1234

Why is the maximum possible value for sinx 1

See the attached diagram.

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

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

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

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

and then there are no solutions?

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

Grandad

**stapel** · April 18th 2009, 08:31 AM

Originally Posted by skeske1234

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

I don't know what you mean by "and then there are no solutions", but I think you mean to ask why sine never takes on values that are larger than 1.

Think back to the definition of the sine. For a right triangle, isn't the sine ratio equal to "(opposite) over (hypotenue)"? For a right triangle, isn't the hypotenuse always the longest of the sides?

Once you get to viewing the sine as a function, rather than a ratio of sides of a right triangle, you can let the angle be ninety degrees (which obviously wouldn't work in a right triangle), and get a value of "1" for the sine.

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Definition of sine and cosine

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