The angle theta is in the fourth quadrant and costheta=frac{2}{7}. Find the exact value of the remaining five trigonometric functions.

The angle theta is in the fourth quadrant and costheta=frac{2}{7}. Find the exact value of the remaining five trigonometric functions.

Question
Trigonometry
asked 2021-02-09
The angle \(\displaystyle\theta\) is in the fourth quadrant and \(\displaystyle{\cos{\theta}}={\frac{{{2}}}{{{7}}}}\). Find the exact value of the remaining five trigonometric functions.

Answers (1)

2021-02-10
It is given that \(\displaystyle\theta\) is in the fourth quadrant and
\(\displaystyle{\cos{{\left(\theta\right)}}}={\frac{{{2}}}{{{7}}}}\)
Since,
\(\displaystyle{\cos{{\left(\theta\right)}}}={\frac{{\text{Base(B)}}}{{\text{Hypotenuse(H)}}}}{\cos{\theta}}\) and \(\displaystyle{\sec{\theta}}\)
\(\displaystyle{\frac{{{B}}}{{{H}}}}\)=\(\displaystyle{\frac{{{2}}}{{{7}}}}\)
Then,
B=2, H=7
Use Pythagoras formula to find the perpindicular P,
\(\displaystyle{H}^{{2}}={B}^{{2}}+{P}^{{2}}\)
\(\displaystyle{P}^{{2}}={H}^{{2}}-{B}^{{2}}\)
\(\displaystyle={7}^{{2}}-{2}^{{2}}\)
\(\displaystyle={49}-{4}\)
\(\displaystyle={45}\)
That is,
\(\displaystyle{P}=\sqrt{{{45}}}\)
All the trigonometric ratios except \(\displaystyle{\cos{\theta}}\) and \(\displaystyle{\sec{\theta}}\) are negative as \(\displaystyle\theta\) is in the fourth quadrant. Then
\(\displaystyle{\sec{\theta}}={\frac{{{1}}}{{{\cos{\theta}}}}}\)
\(\displaystyle={\frac{{{1}}}{{{\frac{{{2}}}{{{7}}}}}}}\)
\(\displaystyle={\frac{{{7}}}{{{2}}}}\)
And \(\displaystyle{\sin{\theta}}=-{\frac{{{P}}}{{{H}}}}=-{\frac{{\sqrt{{{45}}}}}{{{7}}}}\)
\(\displaystyle{\csc{\theta}}=-{\frac{{{H}}}{{{P}}}}=-{\frac{{{7}}}{{\sqrt{{{45}}}}}}\)
\(\displaystyle{\tan{\theta}}=-{\frac{{{P}}}{{{B}}}}=-{\frac{{\sqrt{{{45}}}}}{{{2}}}}\)
\(\displaystyle{\cot{\theta}}=-{\frac{{{B}}}{{{P}}}}=-{\frac{{{2}}}{{\sqrt{{{45}}}}}}\)
Hence, the required value of the remaining trigonometric ratios is
\(\displaystyle{\sec{\theta}}={\frac{{{7}}}{{{2}}}},{\sin{\theta}}=-{\frac{{\sqrt{{{45}}}}}{{{7}}}},{\csc{\theta}}=-{\frac{{{7}}}{{\sqrt{{{45}}}}}},{\tan{\theta}}=-{\frac{{\sqrt{{45}}}}{{{2}}}},{\cot{\theta}}=-{\frac{{{2}}}{{\sqrt{{{45}}}}}}\)
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