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Find cos 2a..If a belongs to (pi, 3pi/2) and cos a = (m-n)/(m+n), where m<n , m=-n...

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dir3ctionn3 | Student, College Freshman | eNoter

Posted June 13, 2011 at 4:14 PM via web

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Find cos 2a..

If a belongs to (pi, 3pi/2) and cos a = (m-n)/(m+n), where m<n , m=-n calculate cos 2a . 

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giorgiana1976 | College Teacher | Valedictorian

Posted June 13, 2011 at 4:40 PM (Answer #2)

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From the enunciation, the angle a is in the third quadrant, where the values for the functions sine and cosine are negative, so the angle 2a will be found in the first quadrant, where the values for sin 2a and cos 2a are positive.

For calculating cos 2a, we'll use the nest formula:

cos 2a = 2(cos a)^2 - 1, where cos a = ( m-n )/( m+n )

cos 2a = 2[( m-n )/( m+n )]^2 - 1

We'll calculate the common denominator of the difference from the right side:

cos 2a = [2(m-n)^2 - (m+n)^2]/(m+n)^2

We'll develop the brackets:

cos 2a = (2m^2-4mn+2n^2-m^2-2mn-n^2)/(m+n)^2

cos 2a = (m^2-6mn+n^2)/(m+n)^2

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted November 26, 2011 at 12:54 AM (Answer #3)

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Assume that 2a=a+a

cos (2a) = cos(a+a)=cos a*cos a - sin a*sin a

cos (2a) = cos^2 a - sin^2 a

You have cos a => sin a = sqrt(1 - cos^2 a)

sin a = sqrt(1 - (m-n)^2/(m+n)^2)

sin a = sqrt[(m+n)^2 - (m-n)^2]/((m+n)^2)

Consider the difference of squares (m+n)^2-(m-n)^2=(m+n-m+n)(m+n+m-n)

Reduce opposite terms:

 (m+n)^2-(m-n)^2=4mn

sin a = sqrt[(4mn)/(m+n)^2]

sin a = -2sqrt(mn)/(m+n)

The value of sine function is negative because the angle "a " is in the third sector of trigonometric circle.

The double of the angle "a" will stay in the first sector of trigonometric circle.

Example: a = 200 degrees => 2a = 400 degrees = 360 degrees + 40 degrees = 40 degrees.

cos (2a) = (m-n)^2/(m+n)^2 - (4mn)/(m+n)^2

cos (2a) = (m^2 - 2mn + n^2 - 4mn)/(m+n)^2

Add similar components

cos (2a) = (m^2 - 2mn + n^2 - 4mn)/(m+n)^2 => cos (2a) = (m^2 - 6mn + n^2)/(m+n)^2

ANSWER:cos (2a) = (m^2 - 6mn + n^2)/(m+n)^2

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