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Julli [10]
1 year ago
9

Find the values of which satisfy the equation. 2 cos (2B+30°) = -√3 in the domain OP ≤B≤360°​

Mathematics
1 answer:
PSYCHO15rus [73]1 year ago
5 0

Answer:

60 and 90

Step-by-step explanation:

2 \cos(2 \beta  + 30)  =  -  \sqrt{3}

\cos(2 \beta  + 30)  =  -  \frac{ \sqrt{3} }{2}

2 \beta  + 30 =  \cos {}^{ - 1} ( \frac{   -   \sqrt{3} }{2} )

First solution.

2 \beta  + 30 = 150

2 \beta  = 120

\beta  = 60

Second solution

2 \beta  + 30 = 210

2 \beta  = 180

\beta  = 90

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Allow me to prove the result: odd numbers come in the form 2n-1, because 2n is always even, and the number immediately before an even number is always odd.

So, if we sum the first N odd numbers, we have

\displaystyle \sum_{i=1}^N 2i-1 = 2\sum_{i=1}^N i - \sum_{i=1}^N 1

The first sum is the sum of all integers from 1 to N, which is N(N+1)/2. We want twice this sum, so we have

\displaystyle 2\sum_{i=1}^N i = 2\cdot\dfrac{N(N+1)}{2}=N(N+1)

The second sum is simply the sum of N ones:

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So, the final result is

\displaystyle \sum_{i=1}^N 2i-1 = 2\sum_{i=1}^N i - \sum_{i=1}^N 1 = N(N+1)-N = N^2+N-N = N^2

which ends the proof.

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