trig questions

hi

i have a few questions

for i) where does the minus in front of the square root come from? is it only there because a root can either be positive or negative (although shouldn't it be +/- sign then?) or does it have to do with something else? such as it being an obtuse angle?

for iii) i don't understand the effect the pi in the brackets would have to make it equal to negative k. can somebody please explain?

thank you :smile:

Original post by ashaxo99

hi

i have a few questions

for i) where does the minus in front of the square root come from? is it only there because a root can either be positive or negative (although shouldn't it be +/- sign then?) or does it have to do with something else? such as it being an obtuse angle?

for iii) i don't understand the effect the pi in the brackets would have to make it equal to negative k. can somebody please explain?

thank you :smile:

Part (i)
An obtuse angle is between

\frac{\pi}{2}

and

\pi

.

If you look at the graph of

y=\cos(x)

, you will see that the curve lies below the

x

axis in this region, and so you need the negative sign.

In brief, if

\theta

is obtuse, then

\cos (\theta ) < 0

, hence the negative sign is required.

Part (iii)
If you look at the graph of

y=\sin(x)

, you will see that it cycles every

2\pi

radians. If you go along

\pi

radians, then the output value is the negative of the original output.
The same is true for

\cos(x)

So basically,

\sin(\theta + \pi) = -\sin(\theta)

and

\cos (\theta + \pi) = -\cos(\theta )

What is

\tan (\theta + \pi)

in terms of

\tan (\theta)

? (Think of the period of the tan function).

(edited 6 years ago)

Reply 2

Physics Enemy

19

Original post by ashaxo99

...

If x^2 = 4, x = (+/-) 2. So cos@ = (+/-) Sqrt[1 - (sin@)^2]

You have to pick a +ve or -ve solution. You know cos@ is -ve for an obtuse angle.

iii) is by defn, so you've not done your homework :tongue:

You can see on a graph sin90 = 1 and sin270 = - 1. More generally if sin@ = k, then sin(@ + Pi) = - k.

This is because sin@ is odd, with rotational symmetry about (@ + Pi/2, 0) i.e) the point in-between.

(edited 6 years ago)

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