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4 years ago

FIND sin0 WHERE 0 IS THE ANGLE SHOWN. GIVE AN EXACT VALUE, NOT A DECIMAL APPROXIMATION. Branliest

Mathematics

1 answer:

expeople1 [14]4 years ago

6 0

Answer:

$\frac{\sqrt{15} }{8}$

Step-by-step explanation:

We require to find the third side of the triangle.

Using Pythagoras' identity

The square on the hypotenuse is equal to the sum of the squares on the other 2 sides.

let x represent the third side, then

x² + 7² = 8², that is

x² + 49 = 64 ( subtract 49 from both sides )

x² = 15 ( take the square root of both sides )

x = $\sqrt{15}$

Thus

sinΘ = $\frac{opposite}{hypotenuse}$ = $\frac{\sqrt{15} }{8}$

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This is one pathway to prove the identity.

Part 1

$\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{1}{\tan(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\cot(\theta) = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{\cos(\theta)}{\sin(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)*\sin(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)(1-\cos(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\$

Part 2

$\frac{\sin^2(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)-\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-(\cos(\theta)-\cos^2(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-\cos(\theta)+\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\$

Part 3

$\frac{\sin^2(\theta)+\cos^2(\theta)-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1}{\sin(\theta)} = \frac{1}{\sin(\theta)} \ \ {\checkmark}\\\\$

As the steps above show, the goal is to get both sides be the same identical expression. You should only work with one side to transform it into the other. In this case, the left side transforms while the right side stays fixed the entire time. The general rule is that you should convert the more complicated expression into a simpler form.

We use other previously established or proven trig identities to work through the steps. For example, I used the pythagorean identity $\sin^2(\theta)+\cos^2(\theta) = 1$ in the second to last step. I broke the steps into three parts to hopefully make it more manageable.

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When you're talking factors, you're talking about some sort of integer; that's because “factors” depends on the concept of divisibility, which are virtually exclusive to integers. When you're talking “greater than”, you're excluding complex numbers (where the concept of ordering doesn't exist) and you're probably assuming positive integers. If you are, then no; no positive integer has factors that are larger than it.

If you go beyond positive numbers, that changes. 0 is an integer, and has every integer, except itself, as factors; since its positive factors are greater than zero, there are factors of zero that are greater than zero. If you extend to include negative numbers, you always have both positive and negative factors; and since all positive integers are greater than all negative integers, all negative integers have factors that are greater than them.

Beyond zero, though, no integer has factors whose magnitudes are greater than its own. And that's a principle that can be extended even to the complex integers

Step-by-step explanation:

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FIND sin0 WHERE 0 IS THE ANGLE SHOWN. GIVE AN EXACT VALUE, NOT A DECIMAL APPROXIMATION. Branliest​

FIND sin0 WHERE 0 IS THE ANGLE SHOWN. GIVE AN EXACT VALUE, NOT A DECIMAL APPROXIMATION. Branliest