Given:
[tex](\sin x)^2+(\cos x)^2=1[/tex]
[tex]\cos \theta = \dfrac{4\sqrt{2}}{7}[/tex]
To find:
The value of [tex]\sin \theta[/tex].
Solution:
We have,
[tex]\cos \theta = \dfrac{4\sqrt{2}}{7}[/tex]
Putting this value in the Pythagorean identity, we get
[tex](\sin \theta)^2+(\cos \theta)^2=1[/tex]
[tex](\sin \theta)^2+\left(\dfrac{4\sqrt{2}}{7}\right)^2=1[/tex]
[tex](\sin \theta)^2+\dfrac{16(2)}{49}=1[/tex]
[tex](\sin \theta)^2+\dfrac{32}{49}=1[/tex]
[tex](\sin \theta)^2=1-\dfrac{32}{49}[/tex]
[tex](\sin \theta)^2=\dfrac{49-32}{49}[/tex]
[tex](\sin \theta)^2=\dfrac{17}{49}[/tex]
Taking square root on both sides, we get
[tex]\sin \theta=\pm \sqrt{\dfrac{17}{49}}[/tex]
[tex]\sin \theta=\pm \dfrac{\sqrt{17}}{7}[/tex]
The given value of [tex]\sin \theta[/tex] is positive. So, [tex]\sin \theta= \dfrac{\sqrt{17}}{7}[/tex].
Therefore, the value of value [tex]\sin \theta[/tex] is [tex]\sin \theta= \dfrac{\sqrt{17}}{7}[/tex].
