please help me! I don't understand these problems at all...

Solve the equation for 0 (less than or equal to) x (less than or equal to) 2pi.

1. sin x = sin (x+2)

2. cos x = cos (x+1)

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- Oct 13th 2008, 05:36 PMlolotrig problem: circular functions of real numbers
please help me! I don't understand these problems at all...

Solve the equation for 0 (less than or equal to) x (less than or equal to) 2pi.

1. sin x = sin (x+2)

2. cos x = cos (x+1) - Oct 13th 2008, 06:16 PMticbol
sinX = sin(X +2)

Normally, it would be X = X +2.

But that leads to nowhere.

So do it by expansion:

sinX = sinXcos(2) +cosXsin(2)

sinX -sinXcos(2) = cosXsin(2)

sinX(1 -cos(2)) = cosXsin(2)

Divide both sides by cosX,

tanX(1 -cos(2)) = sin(2)

tanX = sin(2) / (1 -cos(2))

Using the calculator,

tanX = 0.642096 -------------positive tan, so X is in the 1st or 3rd quadrants.

X = arctan(0.642096) = 0.570796 radian

In the 1st quadrant,

X = 0.570796 radian -----------answer.

In the 3rd quadrant,

X = pi +0.570796 = 3.712389 radians ---------answer.

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You should be able to solve the second equation now.

You should find that X is in the 2nd or 4th quadrants,

and X = 2.641593 or 5.783185 radians. - Oct 15th 2008, 06:03 AMlolo
- Oct 15th 2008, 03:35 PMticbol
You mean the following?

sin(X +1) = sinXcos1 +cosXsin1 ?

You thought that should have been

sin(X +1) = sinX +sin1 ?

The first one is the correct one. That is how it is done with trig functions.

-----------------

Example: Find sin(30deg +60deg).

The correct way:

sin(30deg +60deg)

= sin(30deg)cos(60deg) +cos(30deg)sin(60deg)

= (1/2)(1/2) +(sqrt(3)/2)(sqrt(3)/2)

= 1/4 +3/4

= 1

A wrong way:

sin(30deg +60deg)

= sin(30deg) +sin(60deg)

= 1/2 +sqrt(3)/2

= [1 +sqrt(3)] / 2

Well, of course, sin(30deg +60deg) = sin(90deg) = 1.