Question

"From the formula for sin (2t + t) find sin 3t in terms of sin t."

How do I start with this problem and what should I find at the end? Should I be looking at Ptolemy's identities? (Sum and difference formulas?)

Answer 1

You will find the necessary theory here
http://mathworld.wolfram.com/Multiple-AngleFormulas.html
to get the result
\[\large \sin (2t+t)=\sin 3t=3\sin t-4\sin^{3}t\]

Answer 2

There is a simple derivation of the triple angle formula for sine here:
http://www.sciencehq.com/mathematics/trigonometric-multiple-and-sub-multiple-angle-formulas.html

Answer 3

I have another fun approach :)

You can look at the sine as the imaginary part of the complex form cos(x)+i sin(x),
\[\Large\rm \sin(3t)=\text{Im}\left[\cos(3t)+i \sin(3t)\right]\]And then apply De'moivre's Theorem in reverse,\[\Large\rm \text{Im}\left[\cos(3t)+i \sin(3t)\right]=\text{Im}\left[\cos(t)+i \sin(t)\right]^3\]Expand out the stuff (Imma be lazy and call the cosines c, and sines s),\[\Large\rm \text{Im}\left[c^3+3c^2(is)+3c(is)^2+(is)^3\right]\]Simplify:\[\Large\rm \text{Im}\left[c^3+3ic^2s-3cs^2-is^3\right]\]And then simply pick out the imaginary pieces to get your answer!\[\Large\rm 3c^2s-s^3=3\sin(t)\cos^2(t)-\sin^3(t)\]And then I guess we would have to convert the squared cosine to sines :P

This might seem super long and tedious, but it's actually really helpful when dealing with larger coefficients on your angle, like 11t or 25t

Not so great with 3t though hehe
just a fun approach to think about c:

Answer 4

Thanks a lot for your help!

Answer 5

You're welcome :)

Answer 6

I have another one but this one's pretty straight forward.

All you need to know is sin(a+b)=sina*cosb+sin(b)*cos(a), cos(a+b)=cos(a)*cos(b)-sin(a)*sin(b) and sin(a)^2+sin(b)^2=1.

First express sin(2t)=2*sin(t)*cos(t)
Then express sin(2t+t)=sin(2t)*cos(t) +sin(t) *cos(2t)

Replace sin(2t)=2*sin(t)*cos(t) in sin(2t)*cos(t) +sin(t) *cos(2t) and you have

sin(2t+t)=2*sin(t)*cos(t)^2 +sin(t) *cos(2t)

Express cos(t)^2=1-sin(t)^2 (funamental theorem) thus

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

Express cos(2t)=cos(t)^2-sin(t)^2=1-sin(t)^2 (fundamental theorem) and you have

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