A geometry problem by A Former Brilliant Member
⎩ ⎨ ⎧ tan α = 3 1 α = 6 β
Given the above, find 2 1 ( csc 2 β − 3 sec 2 β ) , if α is acute angle.
Answer up to 4 decimal places.
The answer is 3.1622.
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2 solutions
Great solution! I was quite near it but couldn't quite crack it :/
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Thanks.... @Abhay Kumar Easy and nice problem..... @Mehul Arora No problem...It just need to be worked in the right direction... :-) Next time maybe
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@Mehul Arora I ended up solving a cubic equation for tan 2 β , so you had better progress than me
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@Hung Woei Neoh – LOL, I introduced 1/2 to factorize it as 2 sin 4 Beta or something, and then substitute as 2/3 alpha, But it resulted in a bad looking expression. I didn't think of introducing 1 0 I don't remember though, I just clicked View Solution :P
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@Mehul Arora – I only clicked view solution twice. Both were accidents. With this type of questions, I will try my best to solve it, if I can't turn it into something neat, I'll just use my calculator :P
Typo: The third last line should be 1 0 [ sin 4 β sin ( α − 2 β ) ]
Anyway, wonderful solution! You have great trigo manipulation skills!
I am not able to understand what u did in the third step....
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1 0 1 cos 2 β − 1 0 3 sin 2 β
= sin α cos 2 β − cos α sin 2 β ( ∵ tan α = 3 1 ⟹ sin α = 1 0 1 , cos α = 1 0 3 )
= sin ( α − 2 β ) Use sin A cos B − cos A sin B = sin ( A − B )
got it!!! thanks bro...
Why to introduce √10??? If tan a=1/3 Can be directly sin a/ cos a= 1/3 i.e. sin a =1 and cos a= 3
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Noooo, Notice that cos a can never go above 1. i.e., cos ( α ) ≤ 1 . The introduction of 1 0 was only towards getting the Identity to simplify the expression
Very elegant solution. I had some weird cubic I eventually had to use a calculator for, which I hate doing. I have one question though: Why is 1 0 the correct number to use here? I have read the comments, I understand that introducing 1 0 makes for proper sin and cos values (very good idea by the way). But so does 1 7 or 3 1 . If I would have used either of those or some other number > 3 , the math here would lead to the answer being the number I introduced. So why is 1 0 the correct number to use?
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I just had a thought. If you look at this from the unit circle point of view, then opposite = 1 and adjacent = 3 , but that can't work with a hypotenuse of 1 . So if you say opposite = x 1 and adjacent = x 3 where x > 3 , then tan is still 3 1 and the triangle works. Then use Pythagorean Theorem and solve for x you get x = 1 0 . Is that where it came from?
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Exactly.... :-) .... I just thought using sin A cos B-cos A sin B =sin (A-B) can make it a bit simple and it did... :-)
Given tan(3x) = 1/3
We need to evaluate cosecx-3secx
Write 3 as cot(3x) and take LCM and open sin3x and cos3x by using standard results .
On bit simplification u will get as 2/sin3x
Relevant wiki: Sum and Difference Trigonometric Formulas - Problem Solving
Call the expression R . Use csc 2 β = sin 2 β 1 , sec 2 β = cos 2 β 1 and take LCM.
R = sin 4 β 2 sin 2 β cos 2 β cos 2 β − 3 sin 2 β
= 1 0 [ sin 4 β 1 0 1 cos 2 β − 1 0 3 sin 2 β ]
( tan α = 3 1 ⟹ sin α = 1 0 1 , cos α = 1 0 3 )
= 1 0 [ sin 4 β sin α cos 2 β − cos α sin 2 β ]
= 1 0 [ sin 4 β sin ( α − 2 β ) ] ( ∵ sin A cos B − cos A sin B = sin ( A − B ) )
= 1 0 [ sin 4 β sin 4 β ] ∵ α − 2 β = 4 β
= 1 0 ≈ 3 . 1 6 2 2