X-coordinate of Trigonometric Functions (1 Viewer)

BlueGas

Well-Known Member
Joined
Sep 20, 2014
Messages
2,448
Gender
Male
HSC
N/A
Can someone explain to me the working out of question 9. a)? When sinx = cosx, why did they divide sinx by cosx? Shouldn't they minus cosx to the other side?

 
Last edited:

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
Can someone explain to me the working out of question 9. a)? When sinx = cosx, why did they divide sinx by cosx? Shouldn't they minus cosx to the other side?

They divided both sides by cos x so that they could get the equation tan x = 1, which is easy to solve. Subtracting cos x from both sides wouldn't really help as much with solving the equation.
 

BlueGas

Well-Known Member
Joined
Sep 20, 2014
Messages
2,448
Gender
Male
HSC
N/A
They divided both sides by cos x so that they could get the equation tan x = 1, which is easy to solve. Subtracting cos x from both sides wouldn't really help as much with solving the equation.
Are you actually allowed to do that? Because when it comes to normal algebra you would usually minus to find x.
 

integral95

Well-Known Member
Joined
Dec 16, 2012
Messages
779
Gender
Male
HSC
2013
Are you actually allowed to do that? Because when it comes to normal algebra you would usually minus to find x.
Best to do that when there are squares involved such as


 

braintic

Well-Known Member
Joined
Jan 20, 2011
Messages
2,137
Gender
Undisclosed
HSC
N/A
Are you actually allowed to do that? Because when it comes to normal algebra you would usually minus to find x.
In solving (x-1)² = 2(x-1), you CAN'T divide by (x-1) because x=1 is a zero of BOTH the LHS and RHS, and hence a solution of the equation, so that solution is lost in doing the division.

When solving sinx = cosx, you CAN divide by cosx because there are NO values of x for which sinx and cosx are BOTH zero.
So even though the zeros of cosx will lost by dividing, those zeros won't be solutions of the equation.
 
Last edited:

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top