Use identities to find each exact value. (Do not use a calculator).

cos(-pi/12)

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- Jul 26th 2008, 08:02 AM #1

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- Jul 26th 2008, 08:12 AM #2
Hello,

We know that cos(-x)=cos(x)

Thus cos(-pi/12)=cos(pi/12).

pi/12=(pi/6)/2.

--> cos(-pi/12)=cos((1/2) pi/6)

We know that cos(2x)=2cosē(x)-1. Then cos(x)=2cosē(x/2)-1 --> cos(x/2)=+ or - sqrt[(cos(x)+1)/2]

therefore cos(-pi/12)=+ or - sqrt[(cos(pi/6)+1)/2]

Since -pi/12 is in the fourth quadrant, the cosine is positive.

--------> cos(-pi/12)=sqrt[(cos(pi/6)+1)/2]

(and you should know cos(pi/6))

cos(-pi/12)=sqrt(sqrt(3)+2)/2

- Jul 26th 2008, 08:28 AM #3

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cos(-pi/12) = cos(pi/12)

Why? Because cosine is an even function. Now, let's rewrite pi/12 as:

pi/4 - pi/6

So we got now cos(pi/12) = cos(pi/4 - pi/6). Now, recall the sum and difference identites for cosine:

cos(A + B) = cos(A)cos(B) - sin(A)sin(B)

cos(A - B) = cos(A)cos(B) + sin(A)sin(B)

Let's use it then!

cos(pi/4 - pi/6) = cos(pi/4)cos(pi/6) + sin(pi/4)sin(pi/6)

Now, since they asked you not to use the calculator, you must use the special triangles

1. pi/4 - pi/4 - pi/2 (45-45-90)

2. pi/3 - pi/2 - pi/6 (60-90-30)

cos(pi/4 - pi/6) = ( (sqrt(2)/2) x (sqrt(3)/2) ) + ( (sqrt(2)/2) x 1/2 )

= (sqrt(6)/4) + (sqrt(2)/4)

Final answer: ( sqrt(6) + sqrt(2) )/4

- Jul 26th 2008, 08:31 AM #4
Note that our solutions are equal !

( sqrt(6) + sqrt(2) )/4

sqrt(sqrt(3)+2)/2

See in this post of**earboth**the identity : http://www.mathhelpforum.com/math-help/32751-post3.html (the identity can easily be shown by squaring both sides)

- Jul 26th 2008, 08:53 AM #5