How to prove sin 45 = 1/root 2 and root 2/2 same?

Indranil said:
How to prove sin 45 = 1/root 2 and root 2/2 same thing? I am confused.
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A general rule of arithmetic is \(\displaystyle a * 1 = a.\)

Do you accept that?

Another general rule is

\(\displaystyle b \ne 0 \implies \dfrac{b}{b} = 1.\)

Still following?

Finally, a third rule of arithmetic is

\(\displaystyle c * d = 0 \iff c = 0 \text { or } d = 0.\)

Any problem so far?

\(\displaystyle \text {By definition, } \sqrt{p} * \sqrt{p} = p.\)

\(\displaystyle \therefore p \ne 0 \implies \sqrt{p} \ne 0 \implies \sqrt{2} \ne 0 \text { by third rule.}\)

\(\displaystyle sin(45^o) = \dfrac{1}{\sqrt{2}} \implies sin(45^o) = \dfrac{1}{\sqrt{2}} * 1 \text { by first rule.}\)

\(\displaystyle sin(45^o) = \dfrac{1}{\sqrt{2}} * 1 \implies sin(45^o) = \dfrac{1}{\sqrt{2}} * \dfrac{\sqrt{2}}{\sqrt{2}} \text { by second rule.}\)

\(\displaystyle \therefore sin(45^o) = \dfrac{1 * \sqrt{2}}{\sqrt{2} * \sqrt{2}} = \dfrac{\sqrt{2}}{2}.\)
 
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Indranil said:
How to prove sin 45 = 1/root 2 and root 2/2 same thing? I am confused.
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Have you learned how to "rationalize the denominator"?

Multiply the numerator and denominator of \(\displaystyle \frac{1}{\sqrt{2}}\) by \(\displaystyle \sqrt{2}\), and see what you get.
 
I am going to assume that you know that the degree measure of the angles in any triangle sum to 180 degrees. In a right triangle, one angle is, by definition, 90 degrees so the two acute angles sum to 180- 90= 90 degrees.

To prove that sin(45 (degrees)) is \(\displaystyle \frac{1}{\sqrt{2}}\), consider that if a right triangle has one angle with measure 45 degrees then the other acute angle is 90- 45= 45 degrees also. From that, it follows that the two legs have the same length. Let the two legs have length s. Then by the Pythagorean theorem, the hypotenuse has length, c, satisfying \(\displaystyle c^2= s^2+ s^2= 2s^2\). Divide both sides by 2 and \(\displaystyle c^2\) to get \(\displaystyle \frac{s^2}{c^2}= \frac{1}{2}\). Taking the positive square root of both sides gives \(\displaystyle sin(45)= \frac{s}{c}= \frac{1}{\sqrt{2}}\).

To show that \(\displaystyle \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2}\), multiply both numerator and denominator of \(\displaystyle \frac{1}{\sqrt{2}}\) by \(\displaystyle \sqrt{2}\):
\(\displaystyle \frac{1}{\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}}= \frac{\sqrt{2}}{(\sqrt{2})^2}= \frac{\sqrt{2}}{2}\).
 
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JeffM said:
A general rule of arithmetic is \(\displaystyle a * 1 = a.\)

Do you accept that?

Another general rule is

\(\displaystyle b \ne 0 \implies \dfrac{b}{b} = 1.\)

Still following?

Finally, a third rule of arithmetic is

\(\displaystyle c * d = 0 \iff c = 0 \text { or } d = 0.\)

Any problem so far?

\(\displaystyle \text {By definition, } \sqrt{p} * \sqrt{p} = p.\)

\(\displaystyle \therefore p \ne 0 \implies \sqrt{p} \ne 0 \implies \sqrt{2} \ne 0 \text { by third rule.}\)

\(\displaystyle sin(45^o) = \dfrac{1}{\sqrt{2}} \implies sin(45^o) = \dfrac{1}{\sqrt{2}} * 1 \text { by first rule.}\)

\(\displaystyle sin(45^o) = \dfrac{1}{\sqrt{2}} * 1 \implies sin(45^o) = \dfrac{1}{\sqrt{2}} * \dfrac{\sqrt{2}}{\sqrt{2}} \text { by second rule.}\)

\(\displaystyle \therefore sin(45^o) = \dfrac{1 * \sqrt{2}}{\sqrt{2} * \sqrt{2}} = \dfrac{\sqrt{2}}{2}.\)
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Could you show that how do you get 'root p times root p = p?'
 
Indranil said:
Could you show that how do you get 'root p times root p = p?'
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Please tell us:

What is the definition of a^2?

What is the definition of \(\displaystyle \sqrt{a}\)?

What would you get from \(\displaystyle \sqrt{a} * \sqrt{a}\)?
 
Indranil said:
Could you show that how do you get 'root p times root p = p?'
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It is by definition as stated. You really can't argue about definition. I will try to show you why the definition makes sense.

sqrt(25) is 5 since 5*5 = 25. But another way of writing 5 is sqrt(25). So instead of 5*5 = 25 we can write sqrt(25)*sqrt(25) = 25
sqrt(49) is 7 since 7*7 = 49. But another way of writing 7 is sqrt(49). So instead of 7*7 = 49 we can write sqrt(49)*sqrt(49) = 49

The sqrt(p) is that special number when we multiply by itself to get p. This says sqrt(p)*sqrt(p) = p
 
Indranil said:
… how do you get root p times root p = p?
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We previously asked you to type sqrt() for square roots. Please try to memorize this notation.

You're asking why sqrt(p) ∙ sqrt(p) = p

When a number multiplied by itself equals p, we define such number as a "square root of p".

In other words, sqrt(p) represents a number whose square is p:

sqrt(p) ∙ sqrt(p) = p

I would like to suggest that you study pre-algebra topics, before studying trigonometry. :idea: Students waste a lot of their time, when they approach math studies by jumping back and forth between just parts of basic and advanced topics.
 
Subhotosh Khan said:
Please tell us:

What is the definition of a^2?

What is the definition of \(\displaystyle \sqrt{a}\)?

What would you get from \(\displaystyle \sqrt{a} * \sqrt{a}\)?
Click to expand...
a^2 = a x a
(root) a = a^(1/2)
(root) a x (root) a = a^(1/2) x a^(1/2) = a^1 = a
Am I right?
 
Indranil said:
a^2 = a x a
(root) a = a^(1/2)
(root) a x (root) a = a^(1/2) x a^(1/2) = a^1 = a
Am I right?
Click to expand...
The 2nd definition is wrong. The sqrt(a) is that special number that when you multiply it by itself will give you a. So sqrt(a)*sqrt(a) = a
 
Just multiply by sqrt2/sqrt2

In degrees:

sin45 = 1/sqrt2

(1*sqrt2)/(sqrt2*sqrt2) =

sqrt2/2.

When you multiply by any number divided by itself ( example 1/1 or 2/2 or x/x), you are just multiplying by 1. As any number divided by itself = 1.

Also, sqrtx * sqrtx = x . Because when you multiply the sqrt of a number by itself, the sqrt cancels.
 
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