# Deriving an equation using Trig formulas

• Mar 23rd 2011, 02:44 PM
AliceFisher
Deriving an equation using Trig formulas
I'm fairly confident in using the Pythagorean, Sum, Difference, Double Angle and half Angle formulas. The problem is that I struggle with the mental gymnastics to demonstrate sin(3x) = something horrendous, in truth I spend far too much time pursuing blind alleys that look right for far too long.

I can get as far as using the Sum rule to break the Sin(3x) into say

Sin(2x + x) = sin(2x)cos(x)+cos(2x)sin(x) = (2sin(x)cos(x))cos(x)+(cos^2(x)-sin^2(x))sin(x)

But getting it into the final form always seems a step too far.

Can anyone offer some pointers as to how you would go about solving such problems (the only example I have is an assignment question so it would be dishonest of me to post a variation on it for help).
• Mar 23rd 2011, 03:39 PM
Quacky
Quote:

Originally Posted by AliceFisher
I'm fairly confident in using the Pythagorean, Sum, Difference, Double Angle and half Angle formulas. The problem is that I struggle with the mental gymnastics to demonstrate sin(3x) = something horrendous, in truth I spend far too much time pursuing blind alleys that look right for far too long.

I can get as far as using the Sum rule to break the Sin(3x) into say

Sin(2x + x) = sin(2x)cos(x)+cos(2x)sin(x) = (2sin(x)cos(x))cos(x)+(cos^2(x)-sin^2(x))sin(x)

But getting it into the final form always seems a step too far.

Can anyone offer some pointers as to how you would go about solving such problems (the only example I have is an assignment question so it would be dishonest of me to post a variation on it for help).

You're almost done! Remember that you can rewrite \$\displaystyle Cos^2(x)-Sin^2(x)\$ as \$\displaystyle (1-2Sin^2(x))\$ or \$\displaystyle (2Cos^2(x)-1)\$ - it will help in some situations (I haven't used it here, although I thought it was worth mentioning).

\$\displaystyle (2Sin(x)Cos(x))Cos(x)+(Cos^2(x)-Sin^2(x))Sin(x)\$ .

So multiply through:
\$\displaystyle 2Sin(x)Cos^2(x)+Cos^2(x)Sin(x)-Sin^3(x)\$

Collect like terms:
\$\displaystyle =3Sin(x)Cos^2(x)-Sin^3(x)\$

\$\displaystyle =3Sin(x)(1-Sin^2(x))-Sin^3(x)\$ (see below for explanation of this stage)

Again, multiply through:
\$\displaystyle =3Sin(x)-3Sin^3(x)-Sin^3(x)\$

Again, collect like terms:
\$\displaystyle =3Sin(x)-4Sin^3(x)\$

You'll have to check my working for silly errors, but the key thing to apply when you've reached the stage you have is just \$\displaystyle Cos^2(x)+Sin^2(x)=1\$ and rearrange that to convert everything into one trig. function - I used it at the indicated stage to convert the cosine function into a sine function.