Question

\[\tan(\frac{ 1 }{ 2 }\cos ^{-1}\frac{ \sqrt{5} }{ 3 }) = ?\]

Answer 1

@satellite73

Answer 2

@satellite73 @satellite73 @satellite73 @satellite73 @satellite73 @satellite73 @satellite73 @satellite73 @satellite73 @satellite73 @satellite73 @satellite73 @satellite73 @satellite73

Answer 3

half angle formula for this one

Answer 4

\[\tan(\frac{\theta}{2})=\frac{\sin(\theta)}{1+\cos(\theta)}\] if i remember correctly

Answer 5

since \(\cos^{-1}(\theta)=\frac{\sqrt{5}}{3}\) this is
\[\frac{\sin(\theta)}{1+\frac{\sqrt{5}}{3}}\]

Answer 6

you do not need \(\theta\) you only need \(\cos(\theta)\) and \(\sin(\theta)\)

Answer 7

to find \(\sin(\theta)\) is the same as being asked
"what is \(\sin(\theta)\) if \(\cos(\theta)=\frac{\sqrt{5}}{3}\) ?
we only need the other side of the triangle, which we get by pythagoras

Answer 8

\[?=\sqrt{3^2-\sqrt{5}^2}=\sqrt{9-5}=\sqrt{4}=2\] so
\[\sin(\theta)=\frac{2}{3}\]

Answer 9

penultimate answer is
\[\frac{\frac{2}{3}}{1+\frac{\sqrt{5}}{3}}\] then you can clean this up

Answer 10

@satellite73 my Approach:

cos^-1 sqrt5/3 = t

cos t = sqrt5/3

tan( t/2) = (Sint/2)/cost/2


1-2sin^2 t/2 = cost

2cos^2 t/2 = cos t

Answer 11

Cos t/2 = sqrt(sqrt5 + 3)/sqrt6

Sin t/2 = sqrt(3-sqrt5)/sqrt6

Answer 12

@satellite73 R u there:)

Answer 13

i am here

Answer 14

Can u Check My Solution..

Answer 15

and @satellite73 i am Unable to PM u....Plzz Make it Liable

Answer 16

i don't think your half angle formula for cosine is right, let me check

Answer 17

\[\cos(t)=2\cos^2(\frac{t}{2})-1\] i think

Answer 18

much easier to use the half angle formula for tangent, which contains sine and cosine

Answer 19

Is nt it tan t/2 = (1-cos t) / sin t

Answer 20

either way
it is the same

Answer 21

you can use
\[\frac{1-\cos(t)}{\sin(t)}=\frac{\sin(t)}{1+\cos(t)}\]

Answer 22

it is easy to see that they are equal, cross multiply and you get
\[\sin^2(t)=1-\cos^2(t)\] which we know is true

Answer 23

Yup....thxxxx....