Core 2 - Question

Solve - 2 - 2cos^2 theta = Sin theta for ( 0 < theta < 360 )

How would i do this from start to finish, my exam is tomorrow and i have know idea how to do this.

Reply 1

student_slasher

use rule cos^2 theta = 1-sin^2 theta --> as sin^2 theta+ cos^2 theta=1

2-2cos^2 theta
to
2-2(1-sin^2 theta)
to
2-2+2sin^2 theta
which is
2sin^2 theta

hope this helped :wink:

(edited 10 years ago)

Reply 2

purplerainn

To solve such questions, try and express everything in a single variable.

Make the substitution
cos^2 x =1-sin^2 x

(edited 10 years ago)

Reply 3

Timmy12321

Original post by student_slasher

use rule cos^2 theta = 1-sin^2 theta --> as sin^2 theta+ cos^2 theta=1

2-2cos^2 theta
to
2-2(1-sin^2 theta)
to
2-2+2sin^2 theta
which is
2sin^2 theta

hope this helped :wink:

Thanks, whats do i then do for the 0< theta <360 part ?

Reply 4

Blackfyre

Timmy, good luck bro.

Reply 5

student_slasher

Original post by Timmy12321

Thanks, whats do i then do for the 0< theta <360 part ?

I dont really know what you mean...do you mean you're given an angle of theta or you're given a value for sin theta or sin^2 theta?

all i know is if there is a 2 before sin or sin^2 ; you then halve the value/ angle you're given

if it's a just a sin theta value you do that first
but if it's sin^2 theta you root you're value then halve it and put it into you're calculator to get a value from 0-360

tip- drawing a sin curve always helps looking a the other theta values you're not given

Reply 6

joostan

Original post by student_slasher

Original post by Timmy12321

Thanks, whats do i then do for the 0< theta <360 part ?

Spoiler

Take

\arcsin

of these values and then recall that

\sin(180-x) = \sin(x)

And that it is a repeating function, repeating every 360 degrees.

EDIT: Seriously, what is with the constant negging?
All I'm doing is trying to help out by explaining something and you get negged for it?!