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2 years ago

Teniendo en cuenta que tg 45°=1, deduce el valor de sen 45° y de cos 45° mediante las relaciones fundamentales.

Mathematics

1 answer:

Nana76 [90]2 years ago

6 0

Answer:

the value is 0.707

Step-by-step explanation:

The computation is shown below:

Tan (x) = sen (x) ÷ cos (x)

Tan 45° = sen 45° ÷ cos 45° = 1

Now

Sen^2 (x) + cos^2 (x) = 1

Sen^2 (x) + sin^2 (x) = 1 (As sen (x) = u)

2u^2 = 1

u = √1 ÷ √2

= √2 ÷ 2

= 0.707

Hence, the value is 0.707

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2 years ago

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Step-by-step explanation:

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2 years ago

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On the other hand, the general form is given as follows:

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Equations in Standard Form:

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$\bold{1)} \ x^{2}+y^{2}-4x+12y-20=0 \\ \bold{2)} \ x^{2}+y^{2}+6x-8y-10=0 \\ \bold{3)} \ 3x^{2}+3y^{2}+12x+18y-15=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 3, \ the \ equation \ becomes: \\ x^{2}+y^{2}+4x+6y-5=0 \\ \\ \bold{4)} \ 5x^{2}+5y^{2}-10x+20y-30=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 5, \ the \ equation \ becomes: \\ x^{2}+y^{2}-2x+4y-6=0 \\ \\ \bold{5)} \ 2x^{2}+2y^{2}-24x-16y-8=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 2, \ the \ equation \ becomes: \\ x^{2}+y^{2}-12x-8y-4=0$

$\bold{6)} \ x^{2}+y^{2}+2x-12y$

So let's match each equation:

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Then, its general form is:

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Then, its general form is:

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Then, its general form is:

$x^{2}+y^{2}+4x+6y-5=0$

Third. c) matches 3)

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Then, its general form is: $x^{2}+y^{2}+2x-12y-9=0$

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