Question

if \\( cos \theta = \frac { 7 } { 11 } \\), find the value of \\( sin \theta \\) to 2 decimal places.

Explanation:

Step1: Use Pythagorean theorem

In a right - triangle, by the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\). Here, \(c = 11\), \(a = 7\), and \(b=x\). So \(x=\sqrt{11^{2}-7^{2}}=\sqrt{121 - 49}=\sqrt{72}\).

Step2: Calculate \(\sin\theta\)

We know that \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). The opposite side to \(\theta\) is \(x=\sqrt{72}\) and the hypotenuse is \(11\). So \(\sin\theta=\frac{\sqrt{72}}{11}\approx\frac{8.485}{11}\approx0.77\) (using \(\sqrt{72}\approx8.485\)).

Another way: Using the trigonometric identity \(\sin^{2}\theta+\cos^{2}\theta = 1\). Given \(\cos\theta=\frac{7}{11}\), then \(\sin\theta=\sqrt{1-\cos^{2}\theta}=\sqrt{1 - (\frac{7}{11})^{2}}=\sqrt{\frac{121-49}{121}}=\sqrt{\frac{72}{121}}=\frac{\sqrt{72}}{11}\approx0.77\)

Answer:

\(\frac{\sqrt{72}}{11}\) or \(0.77\)