How do you evaluate $cos (2 arctan (\frac{3}{4}))$ $cos(2Arctan(3/4))$ ?

Question

How do you evaluate $cos (2 arctan (\frac{3}{4}))$ $cos(2Arctan(3/4))$ ?

1 Answer

Impact of this question

12767 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-12T18:32:21

$cos (2 arctan (\frac{3}{4})) = \frac{7}{25} .$ $cos(2arctan(3/4))=7/25.$

Explanation:

Let $arctan (\frac{3}{4}) = θ$ $arctan(3/4)=theta$ , sothat, by defn. of $arctan$ $arctan$ function, $tan θ = \frac{3}{4}$ $tantheta=3/4$ , and, $θ \in (- \frac{π}{2}, \frac{π}{2}) = (- \frac{π}{2}, 0) \cup {0} \cup (0, \frac{π}{2}) .$ $theta in (-pi/2,pi/2)=(-pi/2,0)uu{0}uu(0,pi/2).$

Note that, $tan θ = \frac{3}{4} > 0, θ \notin (- \frac{π}{2}, 0]$ $tantheta=3/4 >0, theta !in (-pi/2,0]$ , but, $θ \in (0, \frac{π}{2}) .$ $theta in (0,pi/2).$

Now, reqd. value $= cos (2 arctan (\frac{3}{4})) = cos 2 θ = \frac{1 - {tan}^{2} θ}{1 + {tan}^{2} θ} = \frac{1 - {(\frac{3}{4})}^{2}}{1 + {(\frac{3}{4})}^{2}} = \frac{16 - 9}{16 + 9} = \frac{7}{25} .$ $=cos(2arctan(3/4))=cos2theta=(1-tan^2theta)/(1+tan^2theta)={1-(3/4)^2}/{1+(3/4)^2}=(16-9)/(16+9)=7/25.$

Science

Math

Humanities

... and beyond

How do you evaluate $cos (2 arctan (\frac{3}{4}))$ $cos(2Arctan(3/4))$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you evaluate cos(2Arctan(3/4))cos(2arctan(34))cos(2Arctan(3/4))?

1 Answer

Explanation:

Related questions

Impact of this question

How do you evaluate $cos (2 arctan (\frac{3}{4}))$ $cos(2Arctan(3/4))$ ?