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### Trig Problems

```
Date: 05/07/99 at 11:18:09
From: Lissy Dietrich
Subject: Trig

1) cos2x = sinx
2) cos2x = cosx
3) sin2x = cos2x
4) sin3x = cos3x
5) sin4x = sin2x
6) cos4x = 1-3cos2x

```

```
Date: 05/07/99 at 15:22:54
From: Doctor Rob
Subject: Re: Trig

Thanks for writing to Ask Dr. Math.

The first step is to write all the sines and cosines of different
angles in terms of some common angle t.  In numbers 1 and 2, t = x.
In numbers 5 and 6, t = 2x. (Skip numbers 3 and 4 for a moment.) Use
the identities

cos(2*t) = 1 - 2*sin^2(t)
cos(2*t) = 2*cos^2(t) - 1
sin(2*t) = 2*sin(t)*cos(t)

That should give you an equation involving sin(t) and/or cos(t).
Bring all terms over to the lefthand side, and factor the resulting
equation. Set each factor equal to zero, and solve for sin(t) or
cos(t). That will tell you the values of sin(t) or cos(t) that make
the equation true. From them, you have to find the values of t, and
then of x.

Here's an example of the same kind:

sin(2*x) = cos(x),
2*sin(x)*cos(x) = cos(x),
2*sin(x)*cos(x) - cos(x) = 0,
cos(x)*[2*sin(x) - 1] = 0,
cos(x) = 0     or    2*sin(x) - 1 = 0,
cos(x) = 0     or    sin(x) = 1/2,
x = (n+1/2)*Pi or x = (2*n+1/3)*Pi or x = (2*n+2/3)*Pi,

where n is any integer. The answers in the interval 0 <= x < 2*Pi are

x = Pi/2, 3*Pi/2, Pi/3, or 2*Pi/3.

Double-check by substituting these back into the equation and seeing
that they do all satisfy it.

Numbers 3 and 4 are a little different. They look like sin(t) =
cos(t). The key here is to divide both sides by the cos(t), to give
the equation tan(t) = 1. Now that should tell you what values of t
will make the equation true, and from that you can figure out the
values of x that work.

If you need more help, write again.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Trigonometry

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