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3 years ago

Can someone please help me with these questions!! Thanks,

Mathematics

1 answer:

fiasKO [112]3 years ago

8 0

6a. 1 - 2sin(x)² - 2cos(x)² = 1 - 2(sin(x)² +cos(x)²) = 1 - 2·1 = -1

6c. tan(x) + sin(x)/cos(x) = tan(x) + tan(x) = 2tan(x)

6e. 3sin(x) + tan(x)cos(x) = 3sin(x) + (sin(x)/cos(x))cos(x) = 3sin(x) +sin(x) = 4sin(x)

6g. 1 - cos(x)²tan(x)² = 1 - cos(x)²·(sin(x)²)/cos(x)²) = 1 -sin(x)² = cos(x)²

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Step-by-step explanation:

I like to start off by simplifying the equation! Then, I use inverse operations to narrow it down until I reach the answer!

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