Question

example 1.9. given that sin 18° ≈ 0.3090, cos 18° ≈ 0.9511, and tan 18° ≈ 0.3249, find the six trigonometric function values of 72°.

Explanation:

Step1: Use co - function identities

We know that \(\sin\theta=\cos(90^{\circ}-\theta)\), \(\cos\theta = \sin(90^{\circ}-\theta)\), \(\tan\theta=\cot(90^{\circ}-\theta)\), \(\cot\theta=\tan(90^{\circ}-\theta)\), \(\sec\theta=\csc(90^{\circ}-\theta)\), \(\csc\theta=\sec(90^{\circ}-\theta)\)

Since \(72^{\circ}=90^{\circ}- 18^{\circ}\)

For \(\sin72^{\circ}\):
\(\sin72^{\circ}=\cos(90^{\circ}-72^{\circ})=\cos18^{\circ}\approx0.9511\)

For \(\cos72^{\circ}\):
\(\cos72^{\circ}=\sin(90^{\circ}-72^{\circ})=\sin18^{\circ}\approx0.3090\)

For \(\tan72^{\circ}\):
\(\tan72^{\circ}=\cot(90^{\circ}-72^{\circ})=\cot18^{\circ}=\frac{1}{\tan18^{\circ}}\approx\frac{1}{0.3249}\approx3.0777\)

For \(\cot72^{\circ}\):
\(\cot72^{\circ}=\tan(90^{\circ}-72^{\circ})=\tan18^{\circ}\approx0.3249\)

For \(\sec72^{\circ}\):
\(\sec72^{\circ}=\csc(90^{\circ}-72^{\circ})=\csc18^{\circ}=\frac{1}{\sin18^{\circ}}\approx\frac{1}{0.3090}\approx3.2361\)

For \(\csc72^{\circ}\):
\(\csc72^{\circ}=\sec(90^{\circ}-72^{\circ})=\sec18^{\circ}=\frac{1}{\cos18^{\circ}}\approx\frac{1}{0.9511}\approx1.0514\)

Answer:

\(\sin72^{\circ}\approx0.9511\), \(\cos72^{\circ}\approx0.3090\), \(\tan72^{\circ}\approx3.0777\), \(\cot72^{\circ}\approx0.3249\), \(\sec72^{\circ}\approx3.2361\), \(\csc72^{\circ}\approx1.0514\)