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

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)