Trigo Sums

**Drdj** · February 12th 2010, 05:37 PM

Dear all,

Could you kindly help me with these 2 sums - Q 18 and Q21 - many, many thanks

**Grandad** · February 14th 2010, 09:50 AM

Hello Drdj

Originally Posted by Drdj

Dear all,

Could you kindly help me with these 2 sums - Q 18 and Q21 - many, many thanks

I think one of the reasons no-one has attempted to answer question 18 is that the diagram is not clear. We can't see the graph of $y = 4\cos3x$ at all. This means that it's not possible to be sure what the value of $k$ is. However, here is some working and a diagram showing two possible solutions. You will have to take it from here.

First $y = 4\cos3x$ crosses the $x$ -axis where:

$\cos3x = 0$

So, working in degrees:

$\Rightarrow 3x = 90, 270, 450, 630, ...$

$\Rightarrow x = 30, 90, 150, 210, ...$

See the black graph in the attached diagram.

So let's suppose first that this graph meets $y= 2\sin x + k$ where $x = 30^o$ . Then:

$2\sin30+k=0$

$\Rightarrow k = -1$

So that's one possible value of $k$ . See the yellow coloured graph.

But perhaps the diagram shows a different position of the graph, and they actually meet at a different point at which the first graph crosses the $x$ -axis; for instance, where $x = 210^o$ . Then we get:

$2\sin210+k = 0$

$\Rightarrow k = 1$

This is the pink graph.

Over to you. Can you sort it out now?

For question 21: multiply top-and-bottom by $(1-\sin x)$ :

$\sqrt{\frac{1-\sin x}{1+\sin x}}= \sqrt{\frac{(1-\sin x)^2}{1-\sin^2 x}}$

$= \sqrt{\frac{(1-\sin x)^2}{\cos^2 x}}$

$=\frac{1-\sin x}{\cos x}$ , if we are given that $x$ is acute, because $\cos x$ is then positive.

$=\sec x - \tan x$

But if $x$ is obtuse, $\sin x > 0$ and $\cos x < 0$ . So $\cos x = -\sqrt{\cos^2x}$ . So we would have to say:

$\sqrt{\frac{(1-\sin x)^2}{\cos^2 x}}=-\frac{1-\sin x}{\cos x}$

$=\tan x - \sec x$

Can you see what happens for other values of $x$ ? If you're not sure, look at the graphs of $\sec x$ and $\tan x$ , and see whereabouts $\sec x >\tan x$ and whereabouts $\tan x > \sec x$ . And then bear in mind that $\sqrt{\frac{1-\sin x}{1+\sin x}}$ has to be positive.

Grandad

**Drdj** · February 14th 2010, 04:44 PM

Dear Grandad,

You're simply wonderful !!

Apologies for the not-so-clear scan of Qn 18 - causing the confusion ....
But thank you so much for your detailed and clear explanation, not to mention to the colorful graphs plotted !!

)

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