The trigonometric functions sec(135^@)

Question
Trigonometric Functions
asked 2021-01-31
The trigonometric functions \(\displaystyle{\sec{{\left({135}^{\circ}\right)}}}\)

Answers (1)

2021-02-01
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
\(\displaystyle\theta={135}^{\circ}={90}^{\circ}+{45}^{\circ}\)
\(\displaystyle{\cos{{\left({135}^{\circ}\right)}}}={\cos{{\left({90}^{\circ}+{45}^{\circ}\right)}}}\)
\(\displaystyle=-{\sin{{\left({45}^{\circ}\right)}}}{\left\lbrace{\cos{{\left({90}^{\circ}+\theta\right)}}}=-{\sin{{\left(\theta\right)}}}\right.}\)
\(\displaystyle=-\frac{{1}}{\sqrt{{2}}}\)
Now \(\displaystyle{\sec{{\left(\theta\right)}}}=\frac{{1}}{{{\cos{{\left(\theta\right)}}}}}\)
\(\displaystyle{\sec{{\left({135}^{\circ}\right)}}}=\frac{{1}}{{{\cos{{\left({135}^{\circ}\right)}}}}}\)
\(\displaystyle=\frac{{1}}{{-{\sin{{\left({45}^{\circ}\right)}}}}}\)
\(\displaystyle=\frac{{1}}{{-\frac{{1}}{\sqrt{{2}}}}}\)
\(\displaystyle=-\sqrt{{2}}\)
=-1.4142
0

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