1.    If acos3Ɵ + 3acosƟsin2Ɵ = m, asin3Ɵ + 3acos2ƟsinƟ = n, prove that (m+n)2/3 + (m-n)2/3 = 2a2/3
asked May 23, 2013 in Trigonometry Answers by anonymous

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2 Answers

answered 5 days ago by Mathical Level 10 User (53,200 points)

If this is an identity it has to be true for all a and theta.

Let ø=90 degrees, then m=0 and n=-a-3a=-4a.

Let ø=30 then m=9a/4 and n=7a/4. 

So in the first case we have to prove 0+4a*2/3=2a*2/3 which is clearly untrue. Even if 2/3 was an exponent, the two sides of the supposed equality are unequal. So instead of an identity it becomes an equation to be solved for theta.

In the second case we have 8a/3+a/3=3a≠2a(2/3).

The question needs to be corrected, perhaps to solve for theta.

The apparent appearance of the fraction 2/3 on each term suggests that it can be removed from the supposed identity (if it is a multiplier) in which case 2m=2a and m=a. This implies that a solution is theta=2nπ where n is an integer.

If 2/3 is an exponent the equation may be trickier to solve.

answered 5 days ago by Rod Top Rated User (415,140 points)
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