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Maths C3 - Trigonometry... Help??

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^ Anyone?? :frown: :frown: :frown:
Original post by Philip-flop
Ok. How do I even do part (c) and (d)?? :frown:

C3 Exe 7A Q12(a).png

For part (c)... Shall I start off by using the fact that...
tanθ=sinθcosθ tan \theta = \frac{sin \theta}{cos \theta} ?

Look back at question 5H that you did successfully a few days back.

You need to use tan⁡45=1\tan 45 = 1 again to write the fraction in the form

tan⁡A+tan⁡B1−tan⁡Atan⁡B\displaystyle \frac{\tan A + \tan B}{1-\tan A \tan B}

For d), use the fact that sin⁡45=cos⁡45=12\displaystyle \sin 45 = \cos 45 = \frac{1}{\sqrt{2}}

These are all quite tricky questions that you don't really see in exams. Don't worry too much if you find them hard.
Original post by notnek
Look back at question 5H that you did successfully a few days back.

You need to use tan⁡45=1\tan 45 = 1 again to write the fraction in the form

tan⁡A+tan⁡B1−tan⁡Atan⁡B\displaystyle \frac{\tan A + \tan B}{1-\tan A \tan B}

For d), use the fact that sin⁡45=cos⁡45=12\displaystyle \sin 45 = \cos 45 = \frac{1}{\sqrt{2}}

These are all quite tricky questions that you don't really see in exams. Don't worry too much if you find them hard.

Yeah I had a feeling that part (c) was similar to Question 5(h) that I did not long ago. Managed to figure that one out now. But am still stuck on part(d). :frown:

Thanks a lot for your help :smile:
(edited 7 years ago)
Original post by Philip-flop
Yeah I had a feeling that part (c) was similar to Question 5(h) that I did not long ago. Managed to figure that one out now. But am still stuck on part(d). :frown:

Thanks a lot for your help :smile:

12(sin⁡θ+cos⁡θ)=12sin⁡θ+12cos⁡θ\displaystyle \frac{1}{\sqrt{2}}\left( \sin \theta + \cos \theta \right) = \frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta

sin45=cos45=12\displaystyle sin 45 = cos 45 = \frac{1}{\sqrt{2}}

Try substituting these to get something of the form sin⁡Acos⁡B+sin⁡Bcos⁡A\sin A \cos B + \sin B \cos A
Original post by notnek
12(sin⁡θ+cos⁡θ)=12sin⁡θ+12cos⁡θ\displaystyle \frac{1}{\sqrt{2}}\left( \sin \theta + \cos \theta \right) = \frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta

sin45=cos45=12\displaystyle sin 45 = cos 45 = \frac{1}{\sqrt{2}}

Try substituting these to get something of the form sin⁡Acos⁡B+sin⁡Bcos⁡A\sin A \cos B + \sin B \cos A


Oh yeah of course! I actually started doing that then realised. So I've done...

12(sinθ+cosθ) \frac{1}{\sqrt 2} (sin \theta + cos \theta)

=12sinθ+12cosθ = \frac{1}{\sqrt 2} sin \theta + \frac{1}{\sqrt 2} cos \theta

=sinθcosπ4+cosθsinπ4 = sin \theta cos \frac{\pi}{4} + cos \theta sin \frac{\pi}{4}

=sin(θ+π4) = sin( \theta + \frac{\pi}{4})

Alternatively I could have done....
12(sinθ+cosθ) \frac{1}{\sqrt 2} (sin \theta + cos \theta)

=12sinθ+12cosθ = \frac{1}{\sqrt 2} sin \theta + \frac{1}{\sqrt 2} cos \theta

=sinπ4sinθ+cosπ4cosθ= sin \frac{\pi}{4} sin \theta + cos \frac{\pi}{4} cos \theta

=cos(π4−θ) = cos (\frac{\pi}{4} - \theta)

Am I right??
Original post by Philip-flop
Oh yeah of course! I actually started doing that then realised. So I've done...

12(sinθ+cosθ) \frac{1}{\sqrt 2} (sin \theta + cos \theta)

=12sinθ+12cosθ = \frac{1}{\sqrt 2} sin \theta + \frac{1}{\sqrt 2} cos \theta

=sinθcosπ4+cosθsinπ4 = sin \theta cos \frac{\pi}{4} + cos \theta sin \frac{\pi}{4}

=sin(θ+π4) = sin( \theta + \frac{\pi}{4})

Alternatively I could have done....
12(sinθ+cosθ) \frac{1}{\sqrt 2} (sin \theta + cos \theta)

=12sinθ+12cosθ = \frac{1}{\sqrt 2} sin \theta + \frac{1}{\sqrt 2} cos \theta

=sinπ4sinθ+cosπ4cosθ= sin \frac{\pi}{4} sin \theta + cos \frac{\pi}{4} cos \theta

=cos(π4−θ) = cos (\frac{\pi}{4} - \theta)

Am I right??

Yes that's all correct.

And another way you can see that those two forms are equivalent is by using

sin⁡A=cos⁡(π2−A)\displaystyle \sin A = \cos \left(\frac{\pi}{2}-A \right) :

sin⁡(θ+π4)=cos⁡[π2−(θ+π4)]=cos⁡(π4−θ)\displaystyle \sin \left( \theta + \frac{\pi}{4} \right) = \cos \left[ \frac{\pi}{2} - \left(\theta + \frac{\pi}{4}\right) \right] = \cos \left(\frac{\pi}{4} - \theta \right)
Original post by notnek
Yes that's all correct.

And another way you can see that those two forms are equivalent is by using

sin⁡A=cos⁡(π2−A)\displaystyle \sin A = \cos \left(\frac{\pi}{2}-A \right) :

sin⁡(θ+π4)=cos⁡[π2−(θ+π4)]=cos⁡(π4−θ)\displaystyle \sin \left( \theta + \frac{\pi}{4} \right) = \cos \left[ \frac{\pi}{2} - \left(\theta + \frac{\pi}{4}\right) \right] = \cos \left(\frac{\pi}{4} - \theta \right)

Oh yeah! Because sin is a translation of cos by 90 degrees (to the right) parallel to the x-axis, correct?

Thanks again! Things are definitely making sense now :smile:
(edited 7 years ago)
Original post by Philip-flop
Oh yeah! Because sin is a translation of cos by 90 degrees (to the right) parallel to the x-axis, correct?

Thanks again! Things are definitely making sense now :smile:

That's basically correct although a shift of cos⁡x\cos x by π2\frac{\pi}{2} to the right gives

cos⁡(x−π2)\cos \left(x-\frac{\pi}{2} \right)

But as you discovered in an older post, cos is an even function so you have

cos⁡(x−π2)=cos⁡[−(x−π2)]=cos⁡(π2−x)\displaystyle \cos \left(x-\frac{\pi}{2}\right) = \cos \left[-\left(x-\frac{\pi}{2}\right) \right] = \cos \left(\frac{\pi}{2}-x\right)
Original post by notnek
That's basically correct although a shift of cos⁡x\cos x by π2\frac{\pi}{2} to the right gives

cos⁡(x−π2)\cos \left(x-\frac{\pi}{2} \right)

But as you discovered in an older post, cos is an even function so you have

cos⁡(x−π2)=cos⁡[−(x−π2)]=cos⁡(π2−x)\displaystyle \cos \left(x-\frac{\pi}{2}\right) = \cos \left[-\left(x-\frac{\pi}{2}\right) \right] = \cos \left(\frac{\pi}{2}-x\right)

Yes, cos is an "even function" due to it's symmetry on either side of the y-axis. Whereas sin is an odd function.
Ok so I'm stuck again already :frown:

C3 EXE7A Q13.png

Have I even been doing this one right?? :frown:
Attachment not found


Then I don't really know where to go from there.
Photo 06-10-2016, 17 26 42.jpg100.9KB
Original post by Philip-flop
Ok so I'm stuck again already :frown:

C3 EXE7A Q13.png

Have I even been doing this one right?? :frown:
Attachment not found


Then I don't really know where to go from there.

You have 3cos⁡θ3\cos \theta on one side of the equation and 3cos⁡θ\sqrt{3}\cos \theta on the other.

You can move them to the same side and combine them.
Original post by notnek
You have 3cos⁡θ3\cos \theta on one side of the equation and 3cos⁡θ\sqrt{3}\cos \theta on the other.

You can move them to the same side and combine them.


Yeah I thought about doing that, so...

(3−3)cosθ=sinθ (3- \sqrt 3) cos\theta = sin \theta

But then I had no idea where to go from there so I just assumed I was wrong :frown:
Original post by Philip-flop
Yeah I thought about doing that, so...

(3−3)cosθ=sinθ (3- \sqrt 3) cos\theta = sin \theta

But then I had no idea where to go from there so I just assumed I was wrong :frown:


Now divide by cosine to get tan and solve it that way.
Original post by RDKGames
Now divide by cosine to get tan and solve it that way.


Omg, I'm actually going to kick myself!! Managed to get there in the end.

tanθ=3−3 tan \theta = 3- \sqrt 3

and then solved it from there to get... θ=51.7,231.7 \theta = 51.7, 231.7

:smile: :smile: :smile:





Thank you @RDKGames and @notnek I'm forever grateful for your help!!
C3 EXE7A Q13.png

For part (c) should I start by doing?...

cos(θ+25)+sin(θ+65)=1 cos (\theta + 25) + sin (\theta +65) = 1

(cosθcos25−sinθsin25)+(sinθcos65+cosθsin65)=tan45 (cos \theta cos 25 - sin \theta sin 25) + (sin \theta cos 65 +cos \theta sin65) = tan 45

(cosθcos25−sinθsin25)+(sinθcos65+cosθsin65)=sin45cos45 (cos \theta cos 25 - sin \theta sin 25) + (sin \theta cos 65 +cos \theta sin65) = \frac{sin45}{cos45}

and then should I times by cos(45)?? Or am I being completely retarded here? :frown: actually have no idea what I'm doing.
Why do I suck at trigonometry? :frown:
(edited 7 years ago)
Original post by Philip-flop
C3 EXE7A Q13.png

For part (c) should I start by doing?...

cos(θ+25)+sin(θ+65)=1 cos (\theta + 25) + sin (\theta +65) = 1

(cosθcos25−sinθsin25)+(sinθcos65+cosθsin65)=tan45 (cos \theta cos 25 - sin \theta sin 25) + (sin \theta cos 65 +cos \theta sin65) = tan 45

(cosθcos25−sinθsin25)+(sinθcos65+cosθsin65)=sin45cos45 (cos \theta cos 25 - sin \theta sin 25) + (sin \theta cos 65 +cos \theta sin65) = \frac{sin45}{cos45}

and then should I times by cos(45)?? Or am I being completely retarded here? :frown: actually have no idea what I'm doing.
Why do I suck at trigonometry? :frown:

Since 25+65=9025+65 = 90 we have

cos⁡65=sin⁡25\cos 65 = \sin 25 and cos⁡25=sin⁡65\cos 25 = \sin 65

Try using this here and look out for angle sum of 90 in future questions - you need it in different types of trig questions.

Also, don't immediately change 11 into tan⁡45\tan 45 unless you think it will be useful. It was useful for a couple of questions that you've done but not here.
Original post by notnek
Since 25+65=9025+65 = 90 we have

cos⁡65=sin⁡25\cos 65 = \sin 25 and cos⁡25=sin⁡65\cos 25 = \sin 65

Try using this here and look out for angle sum of 90 in future questions - you need it in different types of trig questions.

Also, don't immediately change 11 into tan⁡45\tan 45 unless you think it will be useful. It was useful for a couple of questions that you've done but not here.

Ohhhh I see.

I never knew that if... cos(a)andsin(b) cos(a) and sin(b) where a and b sum up to 90 degrees that... cos(a)=sin(b) cos(a) = sin(b)
Again, the book fails to explain any of that!

Ok, so for this question I would do?...

cos(θ+25)+sin(θ+65)=1 cos (\theta + 25) + sin (\theta +65) = 1

(cosθcos25−sinθsin25)+(sinθcos65+cosθsin65)=1 (cos \theta cos 25 - sin \theta sin 25) + (sin \theta cos 65 +cos \theta sin65) = 1

(cosθsin65−sinθcos65)+(sinθcos65+cosθsin65)=1 (cos \theta sin 65 - sin \theta cos 65) + (sin \theta cos 65 +cos \theta sin65) = 1

2(cosθsin65)=1 2(cos \theta sin65) = 1

But then where from there? :/
Original post by Philip-flop
Ohhhh I see.

I never knew that if... cos(a)andsin(b) cos(a) and sin(b) where a and b sum up to 90 degrees that... cos(a)=sin(b) cos(a) = sin(b)
Again, the book fails to explain any of that!

Ok, so for this question I would do?...

cos(θ+25)+sin(θ+65)=1 cos (\theta + 25) + sin (\theta +65) = 1

(cosθcos25−sinθsin25)+(sinθcos65+cosθsin65)=1 (cos \theta cos 25 - sin \theta sin 25) + (sin \theta cos 65 +cos \theta sin65) = 1

(cosθsin65−sinθcos65)+(sinθcos65+cosθsin65)=1 (cos \theta sin 65 - sin \theta cos 65) + (sin \theta cos 65 +cos \theta sin65) = 1

2(cosθsin65)=1 2(cos \theta sin65) = 1

But then where from there? :/

From there sin⁡65\sin 65 is just a number you can work out on your calculator. So you can move it to the other side along with the 22.

The 'adding to 90' rule is just an application of the identity sin⁡x=cos⁡(90−x)\sin x = \cos \left(90-x\right) e.g. where x is 25.
Original post by notnek
From there sin⁡65\sin 65 is just a number you can work out on your calculator. So you can move it to the other side along with the 22.

The 'adding to 90' rule is just an application of the identity sin⁡x=cos⁡(90−x)\sin x = \cos \left(90-x\right) e.g. where x is 25.


Thank you! Not sure why I panicked when I saw...
2(cosθsin65)=1 2(cos \theta sin65) = 1

:colondollar::colondollar::colondollar:
C3 EXE7A Q13.png

for part (e) where do I go from?...

tanθ−11+tanθ=6tanθ \frac{tan \theta -1}{1+tan \theta} =6tan \theta

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