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Fynjy0 [20]
3 years ago
10

Helpmeeeeeeeeeeeeeeeeeeee

Mathematics
1 answer:
ahrayia [7]3 years ago
4 0

Answer:

first question: c. x = 1

6. B. y = -2x - 8

Step-by-step explanation:

first question:

perpendicular means that when it intersects, it makes a right angle.

the equation y = 5 would be parallel to any of the answers with y = ___ because all of those would result in horizontal lines

either of the x = ___ would be perpendicular to y = 5

since it has to pass through the point (1,3) that means that the equation must be x = 1

the equation x = 1 passes through any point beginning with (1 , _)

6.

parallel means that the equations have the same slope. therefore, we can eliminate both A and C

using the given point (-2 , -4),we can plug into the y = mx + b format to solve

m = -2   x = -2   y = -4

-4 = (-2)(-2) + b --> -4 = 4 + b --> b = -8

finally, our equation comes out to be y = -2x - 8

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It has something to do with the<em> </em><u><em>Weierstrass substitution</em></u>, where we have

$\int\, f(\sin(x), \cos(x))dx = \int\, \dfrac{2}{1+t^2}f\left(\dfrac{2t}{1+t^2}, \dfrac{1-t^2}{1+t^2} \right)dt$

First, consider the double angle formula for tangent:

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\tan\left(2 \cdot\dfrac{x}{2}\right)= \dfrac{2\tan(x/2)}{1-\tan^2(x/2)} = \tan(x)=\dfrac{2t}{1-t^2}

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\sin \arctan t = \dfrac{t}{\sqrt{1 + t^2}} \text{ and } \cos \arctan t = \dfrac{1}{\sqrt{1 + t^2}}

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Similarly for cosine, consider the double angle identity

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\cos(x)=  \left(\dfrac{1}{\sqrt{1 + t^2}}\right)^2- \left(\dfrac{t}{\sqrt{1 + t^2}}\right)^2 = \dfrac{1}{t^2+1}-\dfrac{t^2}{t^2+1} =\dfrac{1-t^2}{1+t^2}

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======================================================

5\cos(x) =12\sin(x) +3, x \in [0, 2\pi ]

Solving

5\,\overbrace{\frac{1-t^2}{1+t^2}}^{\cos(x)} = 12\,\overbrace{\frac{2t}{1+t^2}}^{\sin(x)}+3

\implies \dfrac{5-5t^2}{1+t^2}= \dfrac{24t}{1+t^2}+3 \implies  \dfrac{5-5t^2 -24t}{1+t^2}= 3

\implies 5-5t^2-24t=3\left(1+t^2\right) \implies -8t^2-24t+2=0

t = \dfrac{-(-24)\pm \sqrt{(-24)^2-4(-8)\cdot 2}}{2(-8)} = \dfrac{24\pm 8\sqrt{10}}{-16} =  \dfrac{3\pm \sqrt{10}}{-2}

t=-\dfrac{3+\sqrt{10}}{2}\\t=\dfrac{\sqrt{10}-3}{2}

Just note that

\tan\left(\dfrac{x}{2}\right) =  \dfrac{3\pm 8\sqrt{10}}{-2}

and  \tan\left(\dfrac{x}{2}\right) is not defined for x=k\pi , k\in\mathbb{Z}

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