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# Trigonometric Identities watch

1. Simplify:
sin(x - α)cos(α) + cos(x - α)sin(α)

I used the trig identities on sin(x - α) and cos(x - α), and I ended up with:

sin(x - α)cosα = sin(x)cos2(α) - sin(α)cos(x)cos(α)

cos(x - α)sinα = sin(α)cos(x)cos(α) + sin(x)sin2(α)

Im not sure If I've correctly expanded them out, but even if I have, how would I simplify that?
2. (Original post by znx)
Simplify:
sin(x - α)cos(α) + cos(x - α)sin(α)

I used the trig identities on sin(x - α) and cos(x - α), and I ended up with:

sin(x - α)cosα = sin(x)cos2(α) - sin(α)cos(x)cos(α)

cos(x - α)sinα = sin(α)cos(x)cos(α) + sin(x)sin2(α)

Im not sure If I've correctly expanded them out, but even if I have, how would I simplify that?
Do you know that ? What happens if ?
3. (Original post by znx)
Simplify:
sin(x - α)cos(α) + cos(x - α)sin(α)

I used the trig identities on sin(x - α) and cos(x - α), and I ended up with:

sin(x - α)cosα = sin(x)cos2(α) - sin(α)cos(x)cos(α)

cos(x - α)sinα = sin(α)cos(x)cos(α) + sin(x)sin2(α)

Im not sure If I've correctly expanded them out, but even if I have, how would I simplify that?
The above notwithstanding, what you've done is also correct. Note that if you add the two you get

Does a certain famous identity come to mind?
4. Thank you both, I didnt notice that the whole expression was the result of sin(A + B)
5. (Original post by Zacken)
The above notwithstanding, what you've done is also correct. Note that if you add the two you get

Does a certain famous identity come to mind?
Yeah thats the way I done it in the end, however your way was more much more simpler and didn't require expanding each individually.
6. (Original post by znx)
Simplify:
sin(x - α)cos(α) + cos(x - α)sin(α)

I used the trig identities on sin(x - α) and cos(x - α), and I ended up with:

sin(x - α)cosα = sin(x)cos2(α) - sin(α)cos(x)cos(α)

cos(x - α)sinα = sin(α)cos(x)cos(α) + sin(x)sin2(α)

Im not sure If I've correctly expanded them out, but even if I have, how would I simplify that?
7. (Original post by brainmaster)
yes

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