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Temka [501]
4 years ago
8

Please help: please include step by step in how to do this using trig identities!

Mathematics
1 answer:
MakcuM [25]4 years ago
8 0
One would note, that, you never really asked anything above

but.. in case you meant as in proof on the identities

well, doing the left-hand-side   \bf \cfrac{1}{1+sin(x)}\cdot \cfrac{1-sin(x)}{1-sin(x)}\implies \cfrac{1-sin(x)}{[1+sin(x)][1-sin(x)]}\\\\
-----------------------------\\\\
recall\qquad \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
and\qquad sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)\\\\
-----------------------------\\\\
thus
\\\\\\
\cfrac{1-sin(x)}{1^2-sin^2(x)}\implies \cfrac{1-sin(x)}{1-sin^2(x)}\implies \cfrac{1-sin(x)}{cos^2(x)}
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Answer:

1/64

Step-by-step explanation:

Given the expression

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4√4y = 4-3

4√4y = 1

√4y = 1/4

Square both sides

(√4y)²= (1/4)²

4y = 1/16

y = 1/16 * 1/4

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3 years ago
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3 years ago
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Y=4x +5 y=x -4 a (3, 17) b (-3, -7) c (-3, 7) d (3, 7)
svlad2 [7]

Answer:

(-3, -7)

Step-by-step explanation:

y = 4x + 5

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add 4 to both sides of the second equation to get y + 4 = x

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combine like terms y = 4y + 21

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add 4 on both sides to get -3 = x

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Step-by-step explanation:

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Since all triangles r equal to 180° 
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