Question

Trigonometry,

How would you solve this ? part (a)
https://www.dropbox.com/s/39qxunu4r1gfn33/Screenshot%202016-02-28%2017.55.40.png?dl=0

Answer 1

try substituting
x = cos y.

Answer 2

\(\sin [\cos^{-1} (\cos y)] = ... ?\)

Answer 3

or simply,
whats
\(\cos^{-1}\cos y =.. \)

Answer 4

cosy = cos(theta)

Answer 5

theta? where does theta come from?

Answer 6

cosy = cos(x) sorry

Answer 7

how did you get an equation??

we were working on an expression :P

Answer 8

hmm

Answer 9

\(\sin \cos^{-1}x = \sin cos^{-1}\cos y = \sin y = ...\)
makes sense till here?

Answer 10

yes

Answer 11

now we put x =cos y.
can you find sin y from here?

Answer 12

I have sin(cos^-1(cosy))

Answer 13

isn't \(\cos^{-1}\cos y = y\)
?

Answer 14

aah I see

Answer 15

so that equals sin(y)

Answer 16

yes

Answer 17

now we put x =cos y.
can you find sin y from here?

Answer 18

do I have to use any trig identities to tranform it first

Answer 19

you can, there are many ways

Answer 20

sqrt(1 - x)

Answer 21

see if you get \(\sin y = \sqrt{1-x^2}\)

Answer 22

a wild guess

Answer 23

why guess?
\(\sin^2 y +\cos^2 y =1\)

Answer 24

\(\sin^2 y +x^2 =1 \\ \sin^2 y = 1-x^2 \\ \sin y = \sqrt{1-x^2}\)
ok?

Answer 25

I see

Answer 26

so we have sin (cos^-1(x) = sqrt(1 - x^2)

Answer 27

)*

Answer 28

\(\sin \cos^{-1}x = \sin cos^{-1}\cos y = \sin y = \sqrt{1-x^2} \)

\(\sin \cos^{-1}x = \sqrt{1-x^2} \)
yes!

Answer 29

same procedure for others ? :D

Answer 30

exactly! :)

Answer 31

c or d might be lil bit tricky...but same procedure :)
ask me if you get stuck!

Answer 32

for example , is part (b) sqrt(1/x^2 - 1 ) ?

Answer 33

x = cos y

tan y = sin y/ cos y = sqrt(1-x^2)/ x = sqrt (1/x^2-1)

yes!

Answer 34

Oh ok :D, thanks a lot (= !

Answer 35

welcome ^_^