# Trigo Sums

• Feb 12th 2010, 06:37 PM
Drdj
Trigo Sums
Dear all,

Could you kindly help me with these 2 sums - Q 18 and Q21 - many, many thanks :)
• Feb 14th 2010, 10:50 AM
Hello Drdj
Quote:

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.