Maths C3 - Trigonometry... Help??

Thank you!!! How did you know to multiply top and bottom by cos(x)?



Thank you. I never would have known to multiply the top and bottom by cos(x)

Whenever you see a fraction on the top/bottom of another fraction then it's often useful to get rid of it as soon as possible by multiplying top and bottom by the denominator.

E.g.

$\displaystyle \frac{y-x}{\frac{y}{x}-1}$

If you were trying to simplify this then you could get rid of $\frac{y}{x}$ by multiplying top and bottom of the fraction by $x$. This gives

$\displaystyle \frac{y-x}{\frac{y}{x}-1} = \frac{yx-x^2}{y-x}$

@notnek but for the next bit after I get...

$\frac{cos x - sinx}{cos x + sinx}$

how am I meant to know to do $\frac{cos x - sinx}{cos x + sinx} \times \frac{cos x - sinx}{cos x - sinx}$??

(Exe7C Question 6)

For part (b) how do I know which solutions from...
$\theta = \frac{\pi}{24}, \frac{5 \pi}{24}, \frac{13\pi}{24}, \frac{17\pi}{24}$

...satisfy the equation...
$cos2 \theta - sin2 \theta = \frac{1}{\sqrt 2}$ ??

Do I just have to test each "solution" of theta by subbing them into the equation above using my calculator or is there a quicker way?

They all satisfy it, it even says "answers" (plural) in the question so you just list them.

Where the heck do I even start with part (b)i) ??

Where the heck do I even start with part (b)i) ?? I'm not sure I even understand what the question is asking

For the first one if $\tan \frac{\theta }{2}$ we have $\displaystyle \frac{2t}{1+t^2}+\frac{2(1-t^2)}{1+t^2}=1$.
SO just solve the quadratic equation as you normally would.

get $tan\theta$ from the previous part, using $\frac{sin\theta}{cos\theta}$, in terms of $tan\frac{\theta}{2}$

Oh right! It's starting to make sense now. But where have those 't's come from?

The above user has shortened it, so $t = tan\frac{\theta}{2}$ and has done $sin\theta +2cos\theta = 1$ in terms of t.

But the book says its only...
$\theta = \frac{\pi}{24}, \frac{17\pi}{24}$

Which is why I'm confused

For part (a) can someone please explain how this goes from....
$2cos^2 \frac{\theta}{2}$

to this...
$= \frac{1+cos \theta}{2}$ ??

Ah yes, looked at the question and it makes sense. The two extra solutions come from the fact that when you square $\frac{1}{\sqrt{2}}$ you create an answer which can be achieved from squaring $-\frac{1}{\sqrt{2}}$ so two of your solutions satisfy $\sin(2\theta)-\cos(2\theta)=-\frac{1}{\sqrt{2}}$ which is not what we want.

The easiest way around is to plug the values back through the equation and collect the values you need. You can also work out for which angles $\cos2 \theta - \sin2 \theta$ is positive and consider only the answers which fall into those domains.

It doesn't.

$\cos(\theta)=\cos^2(\frac{\theta}{2})-\sin^2(\frac{\theta}{2})=2\cos^2(\frac{\theta}{2})-1 \implies \cos(\theta)+1=2\cos^2(\frac{ \theta }{2} )$

Can someone help explain from where it says "Occurs when $cos( \theta - 22.6) = 1$ ......" <<Where has the number 1 come from?..
I've worked out the rest of the question and can see that the maximum is 13 but I can't seem to understand the last bit. Why is the smallest value of $\theta$ 22.6 ??