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3 years ago

HELP ASAP!!!

Mathematics

1 answer:

Lera25 [3.4K]3 years ago

5 0

$\\ \sf\Rrightarrow y-1=\dfrac{2}{3}(x+1)$

$\\ \sf\Rrightarrow -3-1=\dfrac{2}{3}(a+1)$

$\\ \sf\Rrightarrow -4=\dfrac{-2(a+1)}{3}$

$\\ \sf\Rrightarrow -2(a+1)=-4(3)$

$\\ \sf\Rrightarrow -2(a+1)=-12$

$\\ \sf\Rrightarrow a+1=\dfrac{-12}{-2}$

$\\ \sf\Rrightarrow a+1=6$

$\\ \sf\Rrightarrow a=6-1$

$\\ \sf\Rrightarrow a=5$

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Step-by-step explanation:

$\frac{\cos\left(2x\right)-\cos\left(4x\right)}{\sin\left(2x\right)+\sin\left(4x\right)}$

Apply formula:

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$\sin\left(A\right)+\sin\left(B\right)=2\cdot\sin\left(\frac{A+B}{2}\right)\cdot\sin\left(\frac{A-B}{2}\right)$

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$=\frac{-2\cdot\sin\left(\frac{2x+4x}{2}\right)\cdot\sin\left(\frac{2x-4x}{2}\right)}{2\cdot\sin\left(\frac{2x+4x}{2}\right)\cdot\cos\left(\frac{2x-4x}{2}\right)}$

$=\frac{-\sin\left(\frac{2x-4x}{2}\right)}{\cos\left(\frac{2x-4x}{2}\right)}$

$=\frac{-\sin\left(\frac{-2x}{2}\right)}{\cos\left(\frac{-2x}{2}\right)}$

$=\frac{-\sin\left(-x\right)}{\cos\left(-x\right)}$

$=\frac{-\cdot-\sin\left(x\right)}{\cos\left(x\right)}$

$=\frac{\sin\left(x\right)}{\cos\left(x\right)}$

$=\tan\left(x\right)$

Hence final answer is

$\frac{\cos\left(2x\right)-\cos\left(4x\right)}{\sin\left(2x\right)+\sin\left(4x\right)}=\tan\left(x\right)$

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Answer:

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