Question

find the exact value of cos(2tan^(-1)(20/21))

Answer 1

\[\cos(2x)=2\cos^2(x)-1\] if i recall correctly, so you only need \(\cos(\tan^{-1}(\frac{20}{21}))\)

Answer 2

which means you only need h, which you find by pythagoras as
\[h=\sqrt{20^2+21^2}=\sqrt{841}\] and so \(\cos(\tan^{-1}(x))=\frac{21}{\sqrt{821}}\)

Answer 3

then use the formula above

Answer 4

2cos^2(21/29)-1=cos2x how do you simplify that

Answer 5

oh no hold on

Answer 6

it is not cosine of 21/29
cosine IS 21/29

Answer 7

it is
\[\cos(\tan^{-1}(\frac{20}{21}))=\frac{21}{29}\]

Answer 8

so
\[\cos(2\tan^{-1}(\frac{20}{21})=2\times (\frac{21}{29})^2-1\]

Answer 9

the 21/29 is the output, not the input

Answer 10

so the final answer is 41/841?

Answer 11

it worked! thanks so much. you're a life saver

Answer 12

do you know how to do this problem?

find all the solutions in the interval [0,2pi), rounded to 5 decimal places (enter all answers as a comma separated list)

5cos(2pi)=4

Answer 13

makes no sense since you have no variable there

Answer 14

\[5\cos(\pi x)=4\]
\[\cos(\pi x)=\frac{4}{5}\]
\[\pi x=\cos^{-1}(\frac{4}{5})\] might work if that is where the variable is

Answer 15

correction..

Answer 16

5cos(2THETA)=4