Question

The trigonometric functions sec (225^@)

Trigonometric Functions
ANSWERED
asked 2020-10-23
The trigonometric functions \(\displaystyle{\sec{{\left({225}^{\circ}\right)}}}\)

Answers (1)

2020-10-24

To find the value of the trigonometric functions, first, convert sec function in terms of \(\displaystyle{\cos{{\left(\theta\right)}}}\).
Use the identity \(\displaystyle{\sec{{\left(\theta\right)}}}=\frac{{1}}{{{\cos{{\left(\theta\right)}}}}}\)
We have
\(\displaystyle\theta={225}^{\circ}={180}^{\circ}+{45}^{\circ}\)
\(\displaystyle{\cos{{\left({225}^{\circ}\right)}}}={\cos{{\left({180}^{\circ}+{45}^{\circ}\right)}}}{\left\lbrace{\cos{{\left({180}+\theta\right)}}}=-{\cos{{\left(\theta\right)}}}\right\rbrace}\)
\(\displaystyle=-{\cos{{\left({45}^{\circ}\right)}}}\)
\(\displaystyle=-\frac{{1}}{\sqrt{{2}}}\)
Now \(\displaystyle{\sec{{\left(\theta\right)}}}=\frac{{1}}{{{\cos{{\left(\theta\right)}}}}}\)
\(\displaystyle{\sec{{\left({225}^{\circ}\right)}}}=\frac{{1}}{{{\cos{{\left({225}^{\circ}\right)}}}}}\)
\(\displaystyle=\frac{{1}}{{-\frac{{1}}{\sqrt{{2}}}}}\)
\(\displaystyle=-\sqrt{{2}}\)
\(=-1.4142\)

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