Trigonometry

maughantourig

According to my log tables:

2sinAcosB = sin(A+B)+sin(A-B)

and

2cosAsinB = sin(A+B)-sin(A-B)

So, 2sinAcosB is not equal to 2cosAsinB?

How do I determine which is A and which is B?

If I have sinx *cos2x *2, is it 2sinAcosB or 2cosAsinB?

gubbie

maughantourig wrote: »

According to my log tables:

2sinAcosB = sin(A+B)+sin(A-B)

and

2cosAsinB = sin(A+B)-sin(A-B)

So, 2sinAcosB is not equal to 2cosAsinB?

How do I determine which is A and which is B?

If I have sinx *cos2x *2, is it 2sinAcosB or 2cosAsinB?

Essentially you put the value with the cos in your equation with your value with with cos in your formula and vice versa

sinx*cos2x*2
2sinAcosB
A=x, B=2x
And then solve

If you're worried about which way to solve it, ie is it 2sinxcos2x or 2cos2xsinx don't be, just choose one and fill it in the way the formula tells you. You'll get the following

2sinxcos2x = sin3x + sin (-x). NOTE: sin(-x) = -sin(x)
= sin3x - sinx
2cos2xsinx = sin3x - sinx

maughantourig

If I had taken 2*sinA*cosB as 2cosAsinB instead of 2sinAcosB wouldn't that give a different answer?

gubbie

maughantourig wrote: »

If I had taken 2*sinA*cosB as 2cosAsinB instead of 2sinAcosB wouldn't that give a different answer?

I hope this makes sense:

maughantourig

Got it!
Thanks and btw your writing is immaculate

yournerd

This is paper 2?

maughantourig

Usually, but you do need to know some trig for the calculus on paper 1.

Jackobyte

Arrange it so that A is whichever angle has the larger value.