Suppose Sin(π) = 3/10 And Sec(π) < 0. Find The
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We are given that sin(π) = 3/10 and sec(π) < 0.
From sin(π) = 3/10, we can write cos(π) = β(1 - sinΒ²(π)) = β(1 - (3/10)Β²) = β(1 - 9/100) = β(91/100).
Since sec(π) = 1/cos(π) and sec(π) < 0, it means that cos(π) < 0.
With these values, we can calculate sin(2π) and cos(2π).
First, we'll compute sin(2π): sin(2π) = 2 * sin(π) * cos(π) = 2 * (3/10) * (β91/10).
Now, we'll compute cos(2π): cos(2π) = cosΒ²(π) - sinΒ²(π) = (β91/10)Β² - (3/10)Β².
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