How can I solve $\cos\left(\frac{1}{2}(\arctan\left(\frac{8}{15}\right)\right)$ using the cosine Half Angle Identity?

Question

$\cos\left(\frac{1}{2}(\arctan\left(\frac{8}{15}\right)\right)$ is the same as $\cos(A/2)$ if $A= \arctan\left(\frac{8}{15}\right)$. Using this, how can I evaluate this with the Cosine Half Angle Identity?

I know that $\tan(x)= \frac{8}{15}$, so $\cos x =15$. From here I used the identity: $\sqrt{(1+15)/2}$, but the answer is not $\sqrt{8}$

Answer 1

The cosine half angle formula is a good idea, but you need to get at $\cos A$. Start from $\cos^2A+\sin^2A=1$. Dividing by $\cos^2A$, you get $$ 1+\tan^2A=\frac{1}{\cos^2A} $$ that implies $$ \cos^2A=\frac{1}{1+\tan^2A} $$ Since you know that $\tan A=8/15$, you can deduce $$ \cos^2A=\frac{1}{1+\dfrac{8^2}{15^2}}=\frac{15^2}{289}=\frac{15^2}{17^2} $$ On the other hand, $A$ is an acute angle, because $0<\arctan x<\pi/2$, for $x>0$. Hence $\cos A>0$ and we can deduce $$ \cos A=\frac{15}{17} $$ Now we can apply the half angle formuls: $$ \sqrt{\frac{1+\cos A}{2}}=\sqrt{\frac{1+(15/17)}{2}}=\sqrt{\frac{16}{17}} $$

Answer 2

If $\frac {A}{B}=\tan x=\frac {\sin x}{\cos x}$ then we can write $\sin x=At$ and $\cos x=Bt$ for some $t.$ Then $1=\sin^2x+\cos^2 x=(A^2+B^2)t^2 .$ So $|t|=\frac {1}{\sqrt {A^2+B^2}}.$ So $$|\sin x| = |At|=\frac {|A|}{\sqrt {A^2+B^2}}$$ and $$|cos x|=|Bt|=\frac {|B|}{\sqrt {A^2+B^2}}.$$

The function $\arctan$ is (usually) defined to take values in $[-\pi/2.\pi/2].$

Let $x=\arctan \frac {8}{15}.$ Then $x\in (0,\pi/2)$ so both $\cos x$ and $\cos \frac {x}{2}$ are positive.

With $A=8$ and $B=15$ and $\cos x>0$ we have $\cos x= \frac {15}{\sqrt {8^2+15^2}\;}= \frac {15}{17}.$

Therefore $$\cos \left(\frac {1}{2}\arctan \frac {8}{15}\right)= \cos \frac {1}{2}x=\sqrt {\frac {1+\cos x}{2}}\;=$$ $$=\sqrt { \frac{1+15/17}{2}\;}=\frac {4}{\sqrt {17}\;}.$$