Let 0 ∘ ≤ α , β , γ ≤ 9 0 ∘ be angles such that sin α − cos β = tan γ sin β − cos α = cot γ
and γ 1 , γ 2 , … , γ i are different possible values of γ .Find
i = 1 ∑ n γ i
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That's a very interesting solution to this problem! Thanks for sharing it!
Your solution is awesome! But just sake of variety, if we square and add the given equations, we will get t a n 2 ( γ ) + c o t 2 ( γ ) = − 2 s i n ( α + β ) , which is easy to solve.
A n e a s y s o l u t i o n i s b y a d d i n g L H S a n d R H S , W e g e t s i n ( α ) + s i n ( β ) − c o s ( α ) − c o s ( β ) = t a n ( γ ) + c o t ( γ ) A s i m p l e a p p l i c a t i o n o f A M − G M I n e q . g i v e s s i n ( α ) = s i n ( β ) = 1 c o s ( α ) = c o s ( β ) = 0 a n d t a n ( γ ) = c o t ( γ ) = 1 S o t h e a n s w e r i s 4 5 .
Can you explain how you applied AM-GM? Note that α , β , γ need not be related.
By AM-GM tan(y)+cot(y)>2 But sin(a)+sin(b)-cos(a)-cos(b) cant exceed 2... So sin(a)=sin(b)=1 and cos(a)=cos(b)=0 and tan(y)=cot(y)=1 and the result follows...
{ sin α − cos β = tan γ sin β − cos α = cot γ ⇒ sin α cos γ − cos β cos γ = sin γ ⇒ sin β sin γ − cos α sin γ = cos γ . . . ( 1 ) . . . ( 2 )
( 1 ) + ( 2 ) : sin ( α − γ ) − cos ( β + γ ) = sin γ + cos γ
Equating the sine and cosine terms, we have:
{ α − γ = α 1 8 0 ∘ − β − γ = γ ⇒ α = 2 γ ⇒ β = 1 8 0 ∘ − 2 γ
⇒ ( 1 ) : sin ( 2 γ ) − cos ( 1 8 0 ∘ − 2 γ ) sin ( 2 γ ) + cos ( 2 γ ) 1 + t 2 2 t + 1 + t 2 1 − t 2 2 t + 1 − t 2 t 3 + t 2 − t − 1 ( t − 1 ) ( t + 1 ) 2 ⇒ t = tan γ = tan γ = tan γ = t [ Let t = tan γ ] = t + t 3 = 0 = 0 = { 1 − 1 ⇒ γ = 4 5 ∘ ⇒ 0 ∘ > γ > 9 0 ∘ rejected
We get the same result if we equate sin ( α − γ ) = cos γ and − cos ( β + γ ) = sin γ
You are not allowed to "equating the sine and cosine terms".
Just because sin A + cos B = sin C + cos D , does not imply that A = C and B = D .
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Multiplying both expressions together gives: ( sin α − cos β ) ( sin β − cos α ) = tan γ cot γ = 1
The LHS:
( sin α − cos β ) ( sin β − cos α )
= cos α cos β + sin α sin β − sin α cos α − sin β cos β
= cos ( α − β ) − 2 1 sin 2 α − 2 1 sin 2 β (Using compound angle formula for cosine and double angle formula for sine )
= cos ( α − β ) − 2 1 ( 2 sin ( α + β ) cos ( α − β ) ) (Using a sum-to-product formula )
= cos ( α − β ) − sin ( α + β ) cos ( α − β )
= cos ( α − β ) ( 1 − sin ( α + β ) )
Since both α and β are between 0 ∘ and 9 0 ∘ , ( α − β ) is between − 9 0 ∘ and 9 0 ∘ , so cos ( α − β ) is between 0 and 1 .
Since both α and β are between 0 ∘ and 9 0 ∘ , ( α + β ) is between 0 ∘ and 1 8 0 ∘ , so sin ( α + β ) is between 0 and 1 , and 1 − sin ( α + β ) is also between 0 and 1 .
Two numbers between 0 and 1 multiplied together gives 1 , so they must be both 1 .
Therefore, cos ( α − β ) = 1 and sin ( α + β ) = 0 , which makes α and β either both 0 ∘ or both 9 0 ∘ .
If both are 0 ∘ , tan γ = sin 0 ∘ − cos 0 ∘ = − 1 , so it is rejected.
If both are 9 0 ∘ , tan γ = sin 9 0 ∘ − cos 9 0 ∘ = 1 , so γ = 4 5 ∘ .