Given that sin theta = 1/4, 0
1 answer:
Answer: cos(Θ) = (√15) / 4
Explanation:
The question states:
1) sin(Θ) = 1/4
2) 0 < Θ < π / 2
3) find cos(Θ)
This is how you solve it.
1) Use the fundamental identity (in this part I use α instead of Θ, just for facility of wirting the symbols, but they mean the same for the case).
2) From which you can find:
3) Replace sin(α) with 1/4
=>
=>
4) Given that the angle is in the first quadrant, you know that cosine is positive and the final answer is:
cos(Θ) =
.
And that is the answer.
Explanation:
The question states:
1) sin(Θ) = 1/4
2) 0 < Θ < π / 2
3) find cos(Θ)
This is how you solve it.
1) Use the fundamental identity (in this part I use α instead of Θ, just for facility of wirting the symbols, but they mean the same for the case).
2) From which you can find:
3) Replace sin(α) with 1/4
=>
=>
4) Given that the angle is in the first quadrant, you know that cosine is positive and the final answer is:
cos(Θ) =
And that is the answer.
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All three series converge, so the answer is D.
The common ratios for each sequence are (I) -1/9, (II) -1/10, and (III) -1/3.
Consider a geometric sequence with the first term <em>a</em> and common ratio |<em>r</em>| < 1. Then the <em>n</em>-th partial sum (the sum of the first <em>n</em> terms) of the sequence is
Multiply both sides by <em>r</em> :
Subtract the latter sum from the first, which eliminates all but the first and last terms:
Solve for :
Then as gets arbitrarily large, the term will converge to 0, leaving us with
So the given series converge to
(I) -243/(1 + 1/9) = -2187/10
(II) -1.1/(1 + 1/10) = -1
(III) 27/(1 + 1/3) = 18