Question

I solved cos(26)cos(19)-sin(26)sin(19) and simplified it down to cos(45) which is sqrt(2)/2. Wolfram alpha however does not agree with me. I used the two cosAcosB = .... and sinAsinB = ... Product to Sum formulas. All numbers are degree values. What am I doing wrong?

Answer 1

cos(45) = cos(26 + 19)

use the sum of 2 angles expansion

cos(26+19) = cos(26)sin(19) - sin(26)cos(19)

\[\cos(45) = \frac{1}{\sqrt{2}}\]

Answer 2

what you did I think is

sin(26 + 19) = sin(26)sin(19) + cos(26)cos(19)

looks a bit like the sin expansion rather than the cos expansion

Answer 3

I don't really see what you did differently, I took those same steps to find cos(45) and rationalized it. What I'm wondering is my use of Product to Sum formulas, is it wrong?

Answer 4

I used two main formulas: \[cosAcosB = 1/2[\cos(A-B) + \cos(A+B)]\] and \[sinAsinB = 1/2[\cos(A-B) - \cos(A+B)]\]

Answer 5

wolfram sometimes drops the ball. But see
cos(26)cos(19)-sin(26)sin(19) - 1/sqrt(2)

Answer 6

http://www.wolframalpha.com/input/?i=cos%2826%29cos%2819%29-sin%2826%29sin%2819%29+-+1%2Fsqrt%282%29

Answer 7

and plugged in 26 for A and 19 for B. the cos(A-B) values became cos(7) and cancelled each other out. the cos(A+B) values became cos(45). Since each was multiplied by 1/2 and then added together, both halves came together to leave me with a single whole cos(45) which I found the exact value as being \[\sqrt{2}/2\]

Answer 8

phi: What is the 1/sqrt(2) part at the end for?

Answer 9

to show you get zero. cos(45)= 1/sqrt(2)

Answer 10

well I recognised is straight away as the sum of 2 angles in cos... 26 + 19 so it was simply cos 45

well if you use that formula

then its \[\frac{1}{2}\cos(26 -19) +\frac{1}{2} \cos(26+19) - \frac{1}{2} \cos(26-19) + \frac{1}{2}(26 + 19) = \cos(26 +19)\]

Answer 11

cos(26 + 19) = cos(45)

etc

Answer 12

Also another quick question, when I am using all of these identities, do I have to convert from decimal to radian to use them properly? For example, one I am currently using calls for cos(2 x theta). Theta is the measure of the angle, and it's given as a decimal, should I convert it to radian or leave it as deg?

Answer 13

looks like you got a sign wrong in the 2nd set of terms

Answer 14

That information should be in the question.... I'd leave it as degrees

Answer 15

So I can just do cos(2 x 285), then make it cos(570), I don't have to convert 285 to radian?

Answer 16

And with your comment about the formula I used, are you saying I used it right or what? Thanks for being so patient and helpful btw.

Answer 17

you can use rads or degs.

Answer 18

your derivation looks ok. 1st eq - 2nd equation gives you campbell_st's

Answer 19

based on your formula the terms 1/2cos(A-B) cancel each other leaving

2*(1/2cos(A+B)) = cos (A+B)

Answer 20

The thing I'd say about your formula is that its not one I use... just through practise I've learn to recognise them..... I don't need a head full of formulae

Answer 21

What am I doing wrong?

you are correct. wolfram did the problem numerically.

Answer 22

Gotcha. Thanks all, I'll take a closer look at campbell's formula. Does it have a name?

Answer 23

sum (or difference) of two angles. much more popular than yours!

Answer 24

its called the sum and difference of 2 angles.... in trig... I just know the expaansions... and recognised... 26 + 19 = 45

Answer 25

gotcha, thanks again!