Answer:
[tex]-\dfrac{209}{241}[/tex]
Step-by-step explanation:
Given that:
[tex]tan\theta = \dfrac{15}{4}[/tex]
Also, [tex]\theta[/tex] is in first quadrant.
To find:
[tex]cos2\theta = ?[/tex]
Solution:
Let us have a look at the cosine of twice the angle in terms of tangent of the angle.
Suppose, we are given the value of [tex]tanA[/tex], then the formula can be written as:
[tex]cos2A = \dfrac{1-tan^2A}{1+tan^2A}[/tex]
In terms of [tex]\theta[/tex], we can re-write the formula as:
[tex]cos2\theta = \dfrac{1-tan^2\theta}{1+tan^2\theta}\\\Rightarrow cos2\theta = \dfrac{1-(\frac{15}{4})^2}{1+(\frac{15}{4})^2}\\\Rightarrow cos2\theta = \dfrac{1-\frac{225}{16}}{1+\frac{225}{16}}\\\Rightarrow cos2\theta = \dfrac{\frac{16-225}{16}}{\frac{16+225}{16}}\\\Rightarrow cos2\theta = \bold{-\dfrac{209}{241}}[/tex]
[tex]\theta[/tex] is in first quadrant, but [tex]2\theta[/tex] can be in the second quadrant therefore, we have a negative value of our answer i.e. [tex]cos2\theta[/tex].
Therefore, the answer is:
[tex]-\dfrac{209}{241}[/tex]
