Trig identity question need checking

[JakePearson]

1. cos(2x+y)
= cos(a+b) = sinasinb + cosacosb
= (2x+y) = sin2xsiny + cos2xcosy
= sin2x = 2sinxcosx and cos2x = cos²x-sin²x
= cos(2x+y) = (cos²x-sin²x)cosy - (2sinxcosx and cos2x)siny
(is this correct)

2. evaluate the following exactly (use $$\sqrt{}$$ in your answer where necessary)

cos (5pi / 12)

how do i complete this question ?!?

 

[rock.freak667]

1. cos(2x+y)
= cos(a+b) = sinasinb + cosacosb
= (2x+y) = sin2xsiny + cos2xcosy
= sin2x = 2sinxcosx and cos2x = cos²x-sin²x
= cos(2x+y) = (cos²x-sin²x)cosy - (2sinxcosx and cos2x)siny
(is this correct)

Do you mean sin2xsiny = (2sinxcosx+cos2x)siny ?

also cos(a+b)=cosacosb - sinasinb

2. evaluate the following exactly (use $$\sqrt{}$$ in your answer where necessary)

cos (5pi / 12)

how do i complete this question ?!?

try putting $\frac{5 \pi}{12}$ as two fractions which you know the sine of and the cosine of.

 

[Bohrok]

cos(a+b) = sinasinb + cosacosb is not the correct identity, so the next line is also incorrect (you also forgot something before (2x+y) on the left side of that equation).
The last two lines are correct except for the last one where copied and pasted and cos2x which doesn't belong.

As for evaluating cos(5$\pi$/12), split 5$\pi$/12 into two numbers you can easily evaluate with trigonometric functions. Split them so you have cos(a + b) and use the the identities above to evaluate.

 

[Mark44]

Also, and similar to a comment I made in another thread, don't put in extraneous stuff.
I.e., you have

cos(2x+y)
= cos(a+b) = sinasinb + cosacosb

I understand why you put it there, but if you have to write this down, put it off to the side so that it's not in the flow of your chain of equalities. You don't have to (and shouldn't) write down everything that goes through your mind.