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beks73 [17]
3 years ago
5

PLZ HELP ME In which table does y vary directly with x?

Mathematics
1 answer:
Sergio [31]3 years ago
3 0

Answer:

column D

Step-by-step explanation:

may the answers right

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We will use the sine and cosine of the sum of two angles, the sine and consine of \frac{\pi}{2}, and the relation of the tangent with the sine and cosine:

\sin (\alpha+\beta)=\sin \alpha\cdot\cos\beta + \cos\alpha\cdot\sin\beta

\cos(\alpha+\beta)=\cos\alpha\cdot\cos\beta-\sin\alpha\cdot\sin\beta

\sin\dfrac{\pi}{2}=1,\ \cos\dfrac{\pi}{2}=0

\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha}

If you use those identities, for \alpha=x,\ \beta=\dfrac{\pi}{2}, you get:

\sin\left(x+\dfrac{\pi}{2}\right) = \sin x\cdot\cos\dfrac{\pi}{2} + \cos x\cdot\sin\dfrac{\pi}{2} = \sin x\cdot0 + \cos x \cdot 1 = \cos x

\cos\left(x+\dfrac{\pi}{2}\right) = \cos x \cdot \cos\dfrac{\pi}{2} - \sin x\cdot\sin\dfrac{\pi}{2} = \cos x \cdot 0 - \sin x \cdot 1 = -\sin x

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\tan \left(x+\dfrac{\pi}{2}\right) = \dfrac{\sin\left(x+\dfrac{\pi}{2}\right)}{\cos\left(x+\dfrac{\pi}{2}\right)} = \dfrac{\cos x}{-\sin x} = -\cot x
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