By considering the relationships between the sides of the right angled triangle (hypotonuese of 12 cm) explain why sin x can never be greater than 1?

May 26, 2018

See Below. Explanation:

We know,

$\sin \theta = \text{opposite"/"hypotenuse}$

Now, we know that, In a Triangle, The side opposite to the greater angle is greater than the other.

In a Right Angled Triangle,

The Right Angle is the greatest angle.

So, The opposite side to it must be the greatest.

So, Hypotenuse is the greatest side.

That means, $\text{Opposite " lt " Hypotenuse}$.

So, $\text{Opposite"/"Hypotenuse} < 1$ [It is an proper fraction].

I.e $\sin \theta < 1$.

You can prove it with the Trigonometric Identities too.

We all know, ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$

As, ${\cos}^{2} \theta$ is a square term, it to be real, has to be positive.

So, ${\cos}^{2} \theta > 0$

So, $- {\cos}^{2} \theta < 0$

$\Rightarrow 1 - {\cos}^{2} \theta < 1$

$\Rightarrow {\sin}^{2} \theta < 1$ [From the identity]

$\Rightarrow {\sin}^{2} \theta - 1 < 0$

$\Rightarrow \left(\sin \theta + 1\right) \left(\sin \theta - 1\right) < 0$

Now, Either $\sin \theta < - 1$ or $\sin \theta < 1$.

We know, Sine is a Periodic Function whose value ranges from $- 1$ to $1$. ($- 1 \le \sin x \le 1$).

So, $\sin \theta \cancel{<} - 1$. But, $\sin \theta < 1$.

Hence Proved again.

Hope this helps.

May 27, 2018

If the value of the $S \in$ is $1$ or greater the leg of the triangle is the same as the hypothenuse and there is only a line not a triangle.

Explanation:

The trig functions are based on the Pythagorean Theorem

${A}^{2} + {B}^{2} = {C}^{2}$

$\sin = A$
$\cos = B$
$h y p = C$

so

${\sin}^{2} + {\cos}^{2} = h y {p}^{2}$

In the classic unit trig triangle the Hypothenuse is 1 so

${\sin}^{2} + {\cos}^{2} = {1}^{2}$

This can be illustrated by a $45$ degree right triangle

$S \in = \cos$ so

$\sin 45 = 0.707$
$\cos 45 = 0.707$

${0.707}^{2} + {0.707}^{2} = {1}^{2}$

$0.5 + 0.5 = 1$

$1 = 1$ Classic Pythagorean theorem.

Now if the value of the sin is 1 the value of the cos must be 0

${1}^{2} + {0}^{2} = {1}^{2}$

$1 = 1$

So if the angles of the triangle are such that $\sin = 1 \mathmr{and} \cos = 0$there is no longer a triangle but just a vertical line because the line adjacent is 0

The value of the $S \in$ cannot be more than 1 because the triangle that the trig functions are based on no longer exists.