Trigonometry! #3
A ∈ ( 0 , 9 0 ∘ ) . Let E = ( cos ( 3 A ) − cos ( 9 A ) ) ( sin ( 9 A ) − sin ( 3 A ) ) . The smallest integral value (in degrees) of A for which E is negative equals
This problem is part of the set Trigonometry .
The answer is 16.
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4 solutions
In your second step you could have written 2 sin 6 A sin 3 A 2 cos 6 A cos 3 A = sin 1 2 A sin 6 A < 0
sin x < 0 for x ∈ ( 1 8 0 , 3 6 0 ) .
∴ min A = 1 6 ∵ sin 1 9 2 < 0 ∵ 1 9 2 > 1 8 0 and and 1 9 2 is smallest multiple of 1 2 after 1 8 0 .
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Exactly what I did! Upvoted! This is better and simpler than the original posted solution. :D
By the way, there's a typo. You should have minimum of A as 1 6 , not 1 2 .
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Ya fixed it :)
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@Shivamani Patil – Nope, not fixed yet. The line " 1 9 2 is smallest multiple of 1 6 after 1 8 0 " is incorrect.
It should be smallest multiple of 1 2 .
The appropriate reasoning is that 1 9 2 is the 1 6 th positive multiple of 1 2 and since it's in the ( 1 8 0 , 3 6 0 ) set, sine of this would be negative. We're taking this for the sin ( 1 2 A ) part while sin ( 6 A ) part still remains positive for this minimum value of A and hence our answer is 1 6 .
EDIT: I see that you've fixed it now. Thanks.
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@Prasun Biswas – In hurry it happened now it is correct.
Oh right! Thanks.
I had never thought that a Trigonometry question would get to Lvl 5 !! +1
I think you are the only one who can beat @Sandeep Bhardwaj sir's record of posting questions . Sandeep sir has posted around 260-270 questions , and you have posted 158 in Trig itself !! Way to go man ⌣ ¨
As you've said, A ∈ ( 0 , 2 π . So the only integral value of A in this interval is 1 . I think, you should verify the statement of the question.
The given expression is positive ( considering angles within ( 0 ° , 3 6 0 ° ) ) for θ ϵ ( 1 8 0 ° , 3 6 0 ° ) .
The first term is never negative on the interval from (0,90). Therefore set the second term equal to zero and see this occurs when A=15. Therefore A must be the smallest integer larger than 15 (which is 16). That's the simple way I solved it.
Assume 3A to be B .Then E=(cosB-cos3B)(sin3B-sinB) .
Note that Sin3x=3sinx-4(sin^3)x and cos3x=4(cos^3)x - 3cosx .
Using the above two equations and basic trigo. equations E will simplify to (2sin4B)(sin^2)B .
Second term being positive as its a squared term ,this will imply sin4B<0 .
This implies 4B>180 deg OR B>45 deg .
Therefore 3A>45 deg OR A>15 deg.
Thus first integer A will satisfy is 16 deg.
( cos 3 A − cos 9 A ) ( sin 9 A − sin 3 A ) < 0
( − 2 sin 6 A sin ( − 3 A ) ) ( 2 sin 3 A cos 6 A ) < 0
sin 6 A cos 6 A sin 2 3 A < 0
∴ One of sin 6 A , cos 6 A , sin 3 A is less than zero.
If sin 6 A , sin 3 A is less than zero , then A < 0 .
But 0 < A , producing a contradiction,
Hence cos 6 A < 0 .
Now the cosine of an angle is negative only when it lies in the I I n d or I I I r d quadrant.
So we can say that 6 A > 9 0 ∘ → A > 1 5 ∘
Finally, we require the smallest integral value of A satisfying this inequality, and hence we have A = 1 6 ∘