Name theorem or postulate?

Mathematics

1 answer:

katen-ka-za [31]3 years ago

6 0

Answer:

SAS

Step-by-step explanation:

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Convert the following sum into a product. cosx + cos3x + cos5x + cos7x

kvv77 [185]

Answer:

$4\cos x\cos (4x)\cos (2x)$

Step-by-step explanation:

Given: $cosx + cos3x + cos5x + cos7x$

To convert: the given sum into product

Solution:

Use formula: $\cos A+\cos B=2\cos \left ( \frac{A+B}{2} \right )\cos \left ( \frac{A-B}{2} \right )$

$cosx + cos3x + cos5x + cos7x=2\cos \left ( \frac{x+3x}{2} \right )\cos \left ( \frac{x-3x}{2} \right )+2\cos \left ( \frac{5x+7x}{2} \right )\cos \left ( \frac{5x-7x}{2} \right )\\=2\cos (2x)\cos (-x)+2\cos (6x)\cos (-x)\\=2\cos (2x)\cos (x)+2\cos (6x)\cos (x)\\=2\cos x\left [ \cos (2x)+\cos (6x) \right ]$

$cosx + cos3x + cos5x + cos7x=2\cos \left ( \frac{x+3x}{2} \right )\cos \left ( \frac{x-3x}{2} \right )+2\cos \left ( \frac{5x+7x}{2} \right )\cos \left ( \frac{5x-7x}{2} \right )\\=2\cos x\left [ \cos (2x)+\cos (6x) \right ]\\=2\cos x\left [2 \cos \left ( \frac{2x+6x}{2} \right )\cos \left ( \frac{2x-6x}{2} \right ) \right ]\\=2\cos x\left [ 2\cos (4x) \cos (-2x) \right ]\\=4\cos x\cos (4x)\cos (2x)$