(cos2A−cos2B)2+(sin2A+sin2B)2=4sin2(A+B)

Mathematics

1 answer:

lbvjy [14]2 years ago

8 0

Answer:

${( \cos2A - \cos2B) }^{2} + {( \sin2A + \sin2 B }^{2}\\ = ( - 2\cos2A \cos2B + { \cos {}^{2} 2A } + \cos {}^{2} 2B) + (2 \sin2A \sin2B + \sin {}^{2} 2A + \sin {}^{2} 2B)\\ \\ { = - 2( \sin2A + \sin 2 B - \cos2 A - \cos2B) }\\ \\ { = - 2( - 4 \sin( A + B) \cos(A + B) } \\ \\ = {\tt{double \: angle \: of \: sine}}\\ \\ { = 2 \times 2(2 \sin (A + B) \cos(A + B)) }\\ \\ { = 4 \sin2(A + B)}$

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Shalnov [3]

Using translation concepts, the correct statement is given by:

If function f was translated down 4 units, the f(x) values would be subtracted by 4. A point in the table for the transformed function would be (1,9).

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A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction in it's definition.

A translation down 4 units means that the output values are subtracted by 4, hence the table would be given by:

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garik1379 [7]

<h3>Answer: 658.3</h3>

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Explanation:

The adjacent complementary angle to that 28 degree angle is 62 degrees because 62+28 = 90.

If the reference angle is the upper acute angle 62 degrees, then the y is the opposite leg and the side 350 is the adjacent leg.

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