11M2U Trigonometry Question (1 Viewer)

[Grinchey]

Hi all,

I have been completing some Trigonometry revision for my Mathematics term 3 exam and I am able to complete most of it with ease. However, even though I can solve most trigonometric equations, there is one that has been stumping me (which I have attached, question K). The question for the problem is: Solve each trigonometric for 0 degrees ≤ x ≤ 360 degrees.
Any steps to a solution would be appreciated.

Thanks

 

[andrew12678]

Hi
If you referring to the question: sinx + cos x=1
1. Change sin x to cos(90-x)
2. Use the sum to product rule for cos : cos(x)+cos(y) = 2cos(x+y/2)cos(x-y/2)
3. You will be be able to get rid of one of the cos with exact values and you solve from there.

Hope it helps

 

[Grinchey]

Hmm thanks for your answer but I haven't learnt that method in class yet And also, the answers end up with having tan(x)=-1

 

[integral95]

Hmm thanks for your answer but I haven't learnt that method in class yet And also, the answers end up with having tan(x)=-1

That would imply that


which means that the solutions generated would not be the same, so maybe the question is wrong

 

[braintic]

take cosx to other side and then divide by cosx. thats how you get tanx=-1 because
sinx/cosx = tanx

Ahhh .... no.

 

[braintic]

Andrew12678's method is 3 unit, and they are not rules you are allowed to quote in any case.

Assuming this is meant to be a 2 unit question:

Square both sides: (sinx+cosx)^2 = 1
sin^2 (x) + 2sinxcosx + cos^2 (x) = 1
2sinxcosx = 0 (since sin^2(x) + cos^2(x) = 1)
Either sinx = 0 or cosx=0

Then solve both.

BUT, since you squared, it is likely you will have picked up extra solutions.
So check all of your answer by substitution.

 

[faisalabdul16]

Ahhh .... no.

ahh shit, i just realised it was sinx + cosx = 1.
i misread the question and thought it was equal to 0.

 

[Squar3root]

Andrew12678's method is 3 unit, and they are not rules you are allowed to quote in any case.

Assuming this is meant to be a 2 unit question:

Square both sides: (sinx+cosx)^2 = 1
sin^2 (x) + 2sinxcosx + cos^2 (x) = 1
2sinxcosx = 0 (since sin^2(x) + cos^2(x) = 1)
Either sinx = 0 or cosx=0

Then solve both.

BUT, since you squared, it is likely you will have picked up extra solutions.
So check all of your answer by substitution.

huehuehue

 

[Squar3root]

Is that Chinese?

and people say I'm racist :/

 

[anomalousdecay]

HAHAHAHAHA you just had to SR

Keep on topic though.....

 

[Squar3root]

HAHAHAHAHA you just had to SR

Keep on topic though.....

he's having a go at me. I am racist but he is moreer racist

 

[mreditor16]

braintic, check your PMs please.

 

[Squar3root]

Umm .... why is that racist? I guess I'll have to assume you were joking.
I honestly have no idea what huehuehue means, and I wasn't having a go at anyone.

Relax, its okay to be racist. Obviously you haven't been exposed to the internet enough.

Mods, pls don't ban me. braintic started it

 

[SpiralFlex]

Come on guys keep on topic.