trig inequality qn (1 Viewer)

[Masaken]

"Show that if 0 < x < pi/4, then sin(2x) > 2sin^2(x). Justify your working."

where would i start with this question? i'm so confused, thanks in advance

 

[Masaken]

+ also how to solve ?? i know one of the solutions is pi/4 + kpi, but i don't know how to get the other solution (answers say it's 5pi/2 + kpi, which to me makes no sense?)

 

[cossine]

"Show that if 0 < x < pi/4, then sin(2x) > 2sin^2(x). Justify your working."

where would i start with this question? i'm so confused, thanks in advance

Observe cos(x)>sin(x) for 0 < x < pi/4.

+ also how to solve View attachment 37406?? i know one of the solutions is pi/4 + kpi, but i don't know how to get the other solution (answers say it's 5pi/2 + kpi, which to me makes no sense?)

Try visualising tan x using the unit circle definition

[pi/4, pi/2) union [5pi/4, 3pi/2)

 

[Average Boreduser]

Or graphically it is easily to look at
I'll show you my working in a sec

 

[Masaken]

Try visualising tan x using the unit circle definition

i think i got it, thanks!

[pi/4, pi/2) union [5pi/4, 3pi/2)

i ended up getting that, thanks a lot

 

[Masaken]

Observe cos(x)>sin(x) for 0 < x < pi/4.

graphically or by solving it? i'm not so sure what to do here

 

[5uckerberg]

"Show that if 0 < x < pi/4, then sin(2x) > 2sin^2(x). Justify your working."

where would i start with this question? i'm so confused, thanks in advance

Simple play


Now we use something called the auxiliary angle for






Here, we see that

considering the domain for the trig values it will be clear that it is greater than 0 so therefore,

 

[Masaken]

Simple play


Now we use something called the auxiliary angle for






Here, we see that

considering the domain for the trig values it will be clear that it is greater than 0 so therefore,

oh, i get it now, thanks a lot, i didn't think to use auxiliary

 

[cossine]

graphically or by solving it? i'm not so sure what to do here

you can multiply by 2sin x on both sides. So you end up with

2sin(x)cos(x) > 2sin^2(x)

=> sin(2x) > 2sin^2(x)

 

[5uckerberg]

+ also how to solve View attachment 37406?? i know one of the solutions is pi/4 + kpi, but i don't know how to get the other solution (answers say it's 5pi/2 + kpi, which to me makes no sense?)

It should be less than but greater than or equal to . The other solution is because when the result is undefined or infinity so therefore, the value has to be smaller than and the comes from the fact the tan function repeats every cycle.

 

[Drongoski]

= (+ve) times (+ve)
= +ve (i.e. > 0)

since over the given domain, sin x > 0 and cos x > sin x.