pi/6 + pi/3 does not equal 1.37! Cosine help! urgent!?

Question

my calculus teacher did the following:

cos 1.37 = cos (pi/6 + pi/3)

because the sum of the cosine of pi/6, [which is about .87] and the cosine of pi/3 [which equals .5] = 1.37

but pi/6 + pi/3 is not equal to 1.37!!
it's their COSINES that equal 1.37
he gets the cosine of the sum of the cosines! LOL

i think this is all wrong. you just cant substitute a number for 2 others just cause that number and the other's cosines are equivalent.
Is there a another way to do this exercise?

My final is tomorrow...any help will be appreciated.

Answer 1

pi/6 + pi/3 = 1.57 approximately, NOT 1.37 as you say. Therefore



cos(pi/6+pi/3) = cos(1.57).



However, cos(pi/6+pi/3) DOES NOT EQUAL cos(pi/6) + cos(pi/3).



It is similar to 4+5 = 9, so (4+5)^2 = 9^2. (^2 means squared).

But 4^2 + 5^2 DOES NOT EQUAL 9^2.

Answer 2

I don't think there is anything wrong with



cos 1.37 = cos (pi/6 + pi/3)



because you are still taking the cosine of 1.37 afterward. But maybe I am misunderstanding something...

Answer 3

You are right my little buddy.

The way that you have the question written would be incorrect, If it was written as:

cos 1.37 = cos (pi/6) + cos(pi/3)

Then it would be correct.



You have to remember the order of operations, BEDMAS. Meaning what's in the brackets always come first.

Answer 4

Cos(A+B)=CosACosB-SinASinB

That's the identity he should be using



Cos(pi/6+pi/3)= Cos(pi/6)*Cos(pi/3) - Sin(pi/6)*Sin(pi/3)

=Cos(30)*Cos(60) - Sin(30)Sin(60)

=.866*.5 - .5*.866

=.433-.433

=0

Look at it this way:

Cos(pi/6+pi/3)_

=Cos(pi/6+2pi/6)

=Cos3pi/6

=Cos pi/2

=Cos90

=0

You are absolutely correct in saying that pi/6+pi/3 does not= 1.87

pi/6=3.1416/6=0.5236

pi/3=3.1416/3=1.0472

.........................--------------

Totapi/6+pi/3=1.5708

Answer 5

Remember that the max value for cossin or sin is 1.

pi/6+pi/3 = pi/2 ~ 1.57



cos(a+b)= cos(a)*cos(b)-sin(a)*sin(b)

cos(a-b)= cos(a)*cos(b)+sin(a)*sin(b)



cos(pi/6+pi/3) = cos(pi/6)*cos(pi/3)-sin(pi/6)*sin(pi/3)



but pi/6 is the complement of pi/3 so we know that:

cos(pi/6)=sin(pi/3) and cos(pi/3)=sin(pi/6)



so we have:

cos(pi/6+pi/3)=sin(pi/3)*sin(pi/6)-sin(pi/6)*sin(pi/3)=0



or....pi/6+pi/3 = pi/2 and cos(pi/2)=0



Also:



cos(a)+cos(b)=2*cos[(a+b)/2]*cos[(a-b)/2]



So we have that cos(pi/3)+cos(pi/6)=2*cos(pi/4)*cos(pi/12)



we know that:

cos(pi/12+pi/12)=cos(pi/12)*cos(pi/12)-sin(pi/12)*sin(pi/12)

= cos(pi/6)=cos^2(pi/12)-[1-cos^2(pi/12)]

=sqrt(3)/2=2*cos^2(pi/12)-1

=cos(pi/12)=sqrt{[1+sqrt(3)/2]/2}



finally, cos(pi/3)+cos(pi/6)=2*cos(pi/4)*sqrt{[1+sqrt(3)/2]/2}

= 2*sqrt(2)/2*sqrt{[1+sqrt(3)/2]/2} ~1.37

Answer 6

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