Help (1 Viewer)

[Kutay]

question

sin2(theta) = tan(theta), for the domain 0 <(orequal) theta >(or equal) PI

got to find all angles i am not sure how to get this what i have done so far is that i have draw the two graphs
circled the point of intersection
not too sure what excatly to do next plz help thanks

 

[acmilan]

pretend the angle is a instead of theta

sin2a = tana
2sinacosa = sina/cosa
2sinacos²a - sina = 0
sina(2cos² - 1) = 0

sina = 0 or 2cos² - 1 = 0

Just solve them now like ordinary 2 unit trig equations

(remember that cosa cant = 0 because it is the denominator of tana)

 

[Trev]

(remember that cosa cant = 0 because it is the denominator of tana)

dangt, would never have thought of that, i don't think we have ever been told that either! thanx acmilan

 

[acmilan]

It doesnt really matter in this question. If you have time in tests you can test your answers in the original equation to see if it works

 

[Trev]

I've never come across a question that has discluded an answer because of this, have you?
Just like I have never thought about the 'holes' formed from, ie. P(x,y) is point perpendicular to (a,b) (c,d), find the equation of P(x,y) - or whater, I hope you get what I mean. Like, how you use the gradient equalling negative reciprocal of the other in the case given, and the denominator not equalling zero......have you come across this in an answer?
please understand my ramblings

 

[acmilan]

I have come across heaps of questions where you had to discard an answer that seems right but isnt when you take a look at the original question

 

[Trev]

Have you in locus questions such as that? As I remember Slide Rule mentioning the holes in such a question one time, that is how I knew of it.

 

[acmilan]

I havnt come across locus ones like that in tests, but definitely trig ones

 

[Trev]

Well thanx again for that, great help!

 

[Kutay]

answer is like sin theta = o
and sin^2 theta = root -1, which you state (no solution)!!!!

 

[Trev]

pretend the angle is a instead of theta

sin2a = tana
2sinacosa = sina/cosa
2sinacos²a - sina = 0
sina(2cos² - 1) = 0

sina = 0 or 2cos² - 1 = 0

Just solve them now like ordinary 2 unit trig equations

(remember that cosa cant = 0 because it is the denominator of tana)

Yeah, so a(or theta)
= 0°, 180° (from sina=0); 45°, 315° (from cos = (1/2)^1/2)