Question

Solve : arctan {1/(cos2@)^1/2} - arctan {(cos2@)^1/2}

Answer 1

@myininaya

Answer 2

I'm going to use x instead of @

does solve mean simplify because I see no equation to actually solve
\[\arctan(\frac{1}{(\cos(2x))^\frac{1}{2}})- \arctan((\cos(2x))^\frac{1}{2})\]

Answer 3

Actually, u = given expression and I need to find 'sin u'

Answer 4

oh
so we want to use difference identity for sine

Answer 5

\[\sin(g-h)=\sin(g)\cos(h)-\sin(h) \cos(g) \\ \text{ where } g=\arctan(\frac{1}{(\cos(2x))^\frac{1}{2}}) \\ \text{ and } h=\arctan((\cos(2x))^\frac{1}{2})\]

Answer 6

we can write sin(g) , cos(h), sin(h), and cos(g) as algebraic expressions

Answer 7

let me know if you have questions so far
I will give you an example of the algebraic expression if you need

Answer 8

I was unable to find the hypotaneous

Answer 9

for the algebraic expression part?

Answer 10

let's look at sin(g)....

Answer 11

when I say sin(g) this is short for
\[\sin(\arctan(\frac{1}{(\cos(2x))^\frac{1}{2}}))\]

Answer 12

Yeah .. I simplified the given expression and got tan u but was unable to get sin u

Answer 13

\[g=\arctan(\frac{1}{(\cos(2x))^\frac{1}{2}}) \\ \tan(g)=\frac{1}{(\cos(2x))^\frac{1}{2}} =\frac{opp}{adj}\]