Question

how to solve cos x/2 sinx/2 and tan x/2 see attached. What did I do wrong?

Answer 1

These use the half angle identity. \[\sin \frac{ \Theta }{ 2 } = \sqrt{\frac{ 1 - \cos \Theta }{ 2 }}\]

and

\[\cos \frac{ \Theta }{ 2 } = \sqrt{\frac{ 1 + \cos \Theta }{ 2 }}\]

Answer 2

What did you use ?

Answer 3

The last of for tan(x/2) is wrong because the identity is \[\tan \frac{ \Theta }{ 2 } = \sqrt{\frac{ 1 - \cos \Theta }{ 1 + \cos \Theta }}\]

Answer 4

\[\csc(x) = \frac{ 1 }{ \sin(x) }\]

so \[\sin(x) = \frac{ 1 }{ 9 }\]

remember sin(x) = opposite/ hypo

Answer 5

so cos(x) = adjacent/ hypo

which is \[\cos(x) = \frac{ \sqrt{80} }{ 9 }\]

Answer 6

Why did you use \[\frac{ -\sqrt{48} }{ 9 }\]

Answer 7

do them all again and note the triangle I used.

Answer 8

there a few things wrong. the formula
\[ \sin \frac{ \Theta }{ 2 } = \sqrt{\frac{ 1 - \cos \Theta }{ 2 }}\]
has the 2 inside the square root on the right side
your 2 is outside the square root

also, the sin(x)= 1/9 and (because x is between 90 and 180)
cos(x) = -sqrt(80)/9

notice that cos(x) is \( -\sqrt{80} / 9 \)
you DO NOT do cos(sqrt(80)/9)

Answer 9

so you should find
\[ \sin\left(\frac{x}{2}\right)= \sqrt{ \frac{1 - \frac{-\sqrt{80}}{9}}{2}} \]

Answer 10

that turns into
\[ \sin\left(\frac{x}{2}\right)= \sqrt{ \frac{1 + \frac{\sqrt{80}}{9}}{2}} \]