Question

cos ( pi/4 + sin^-1 sqrtx/2 ) in terms of x

Answer 1

\[\cos ( \frac{ \pi }{ 4 } =\sin^{-1} \frac{ \sqrt{x} }{ 2 } ) \]

Answer 2

something like that , the question

Answer 3

it should be + symbol not =

Answer 4

\(\large \cos(A+B)=\cos A\cos B-\sin A\sin B\)
so what will be
\(\cos(\pi/4 +y)=....?\)
where, A = pi/4 , B = y = \(\sin^{-1}(\sqrt x/2)\)

Answer 5

but , later on the cos will get abt 0.7071

Answer 6

cos pi/4 = sin pi/4 = 1/sqrt 2
yes.

Answer 7

if cos 45 = 0.7071

Answer 8

okay , i get it now

Answer 9

so the answer will be \[\frac{ \sqrt{(4-x)} }{ 2.\sqrt{2} } - \frac{ \sqrt{x} }{ 2.\sqrt{2} }\]

Answer 10

right ?

Answer 11

how?

Answer 12

err ,from the \[\sin^{-1} \frac{ \sqrt{x} }{ 2 }\] , i find the triangle

Answer 13

ok, yes!
you're correct :)

Answer 14

just checking whether you actually solved it :)

Answer 15

ok ok thank you

Answer 16

welcome ^_^

Answer 17

erm , here's another question , i just want to make sure what i'm doing is right

Answer 18

\[\sec ( \cot ^ {-1} 2x - \frac{ \pi }{ 2 }\]

Answer 19

should we use the cofunction identities ?

Answer 20

@hartnn

Answer 21

yes!
sec (A-pi/2) = csc A

Answer 22

but the question want in terms of x , still need to use that formula ?

Answer 23

yes,
so your Q becomes
csc(cot inverse 2x)

Answer 24

okay , i just know until that , after that what should we do ?

Answer 25

cot inverse 2x = y
so you need csc y
make a triangle...

Answer 26

cot y = 2x

Answer 27

this is the triangle

Answer 28

absolutely correct
just find csc from that

Answer 29

how abt the pi ? should we just leave it like that ?

Answer 30

we already dealt with pi ?
sec (cot inv 2x - pi/2)
= csc (cot inv 2x)

Answer 31

oh okay , i forgot abt that heheh

Answer 32

lo the last answer is \[\sqrt{(1+4x^{2})}\]

Answer 33

i mean so

Answer 34

yes! correct :)

Answer 35

okay , thank you again , if i need you i'll message you :)

Answer 36

sure :)