If cos theta = 3/5 and theta is in the first quadrant, then [tex]\sin 2 \theta=\frac{24}{25}[/tex]
Given: cosθ = 3/5 and θ is in first quadrant
We need to find sin2θ
[tex]\text {We know, } \sin ^{2} \theta=1-\cos ^{2} \theta[/tex]
Substitute cosθ in above formula
[tex]\sin ^{2} \theta=1-\left(\frac{3}{5}\right)^{2}[/tex]
[tex]\sin ^{2} \theta=1-\left(\frac{9}{25}\right)[/tex]
[tex]\begin{array}{l}{\sin ^{2} \theta=\frac{25-9}{25}} \\\\ {\sin ^{2} \theta=\frac{16}{25}}\end{array}[/tex]
Taking square root on both sides,
[tex]\sin \theta=\pm \frac{4}{5}[/tex]
Since, θ is in first quadrant therefore we will choose positive value
[tex]\sin \theta=\frac{4}{5}[/tex]
Let us calculate sin2θ
The formula for sin2θ is given as:
[tex]\sin 2 \theta=2 \sin \theta \cos \theta[/tex]
[tex]\sin 2 \theta=2 \times \frac{3}{5} \times \frac{4}{5}=\frac{24}{25}[/tex]
Thus the value of sin 2 theta is found
