lupitasgirl8890 lupitasgirl8890
  • 25-09-2019
  • Mathematics
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Enter the expression 2cos2(θ)−1 , where θ is the lowercase Greek letter theta. 2cos2(θ)−1 2 c o s 2 ( θ ) − 1 = nothing

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Аноним Аноним
  • 25-09-2019

We have to evaluate

=2 cos 2Ф - 1

⇒cos 2Ф

A =2cos²Ф-1

 B =1 -2 sin²Ф

 C = cos²Ф - sin²Ф

Possible Answers are

1.

=2 ×(2cos²Ф-1)-1

=4cos²Ф-2-1

=4cos²Ф - 3

2.

=2×(1 -2 sin²Ф)-1

=2- 4sin²Ф-1

=1 - 4sin²Ф

3.

=2× ( cos²Ф - sin²Ф) -1

=2 cos²Ф - 2sin²Ф-1

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