Michael's trigonometric expression

Algebra Level 3

Find the smallest positive integer x x such that tan ⁡ x ∘ = cos ⁡ 201 3 ∘ + sin ⁡ 201 3 ∘ cos ⁡ 201 3 ∘ − sin ⁡ 201 3 ∘ . \tan{x^{\circ}}=\dfrac{\cos{2013^{\circ}}+\sin{2013^{\circ}}}{\cos{2013^{\circ}}-\sin{2013^{\circ}}}.

This problem is shared by Michael T. from AIME.


The answer is 78.

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Mei Li by Mei Li

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7 solutions

Adam Dai
Nov 11, 2013

Dividing the numerator and denominator by cos ⁡ 2013 \cos 2013 gives us 1 + tan ⁡ 2013 1 − tan ⁡ 2013 \frac{1 + \tan 2013}{1 - \tan 2013} . If we use the tangent summation formula tan ⁡ ( a + b ) = tan ⁡ a + tan ⁡ b 1 − tan ⁡ a tan ⁡ b \tan(a+b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} with a = 45 a = 45 and b = 2013 b = 2013 , we get tan ⁡ ( 45 + 2013 ) = 1 + tan ⁡ 2013 1 − tan ⁡ 2013 \tan(45 + 2013) = \frac{1 + \tan 2013}{1 - \tan 2013} . Thus, our answer is 45 + 2013 = 2058 45 +2013 = 2058 which is equivalent to 78 \boxed{78} because tangent has a period of π \pi .

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Nice and easy way to solve this question! Voted up!

Fan Zhang - 7 years, 6 months ago

gr8 dude,,,,,,,,

Saravanan Rajenderan - 7 years, 6 months ago
Aj Sedo
Nov 11, 2013

tan x = cos(2013) + sin(2013)/cos(2013) - sin(2013)

tan x = (-0.8387 + -0.5446)/(-0.8387 - (-0.5446))

tan x = -1.3833/-0.2941

tan x = 4.70350221

x = arctan(4.70350221)

x = 78

^_^

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This was an AIME problem, so no calculators :(

Michael Tang - 7 years, 6 months ago
Samuel Hatin
Nov 10, 2013

This problem can be solved with 3 trigonometric identities. Let's start by simplifying the angle's value. How many full rotation of 360 degrees are done in 2013? 5. So we can say that:

tan ⁡ 2013 ≡ tan ⁡ 213 \tan 2013 \equiv \tan 213

Then maybe there is a way to simplify the those cos and sin with the tank so we obtain only sine and cosine. Let's start by removing the denominator and using the definition of tan(x). We have :

sin ⁡ x cos ⁡ 213 cos ⁡ x − sin ⁡ x sin ⁡ 213 cos ⁡ x = cos ⁡ 213 + sin ⁡ 213 \frac{{\sin x\cos 213}}{{\cos x}} - \frac{{\sin x\sin 213}}{{\cos x}} = \cos 213 + \sin 213

Moving cos(x) on the RHS and using these 2 identities we have :

sin ⁡ a sin ⁡ b = 1 2 ( cos ⁡ ( a − b ) − cos ⁡ ( a + b ) ) cos ⁡ a cos ⁡ b = 1 2 ( cos ⁡ ( a − b ) + cos ⁡ ( a + b ) ) sin ⁡ x cos ⁡ 213 − sin ⁡ x sin ⁡ 213 = cos ⁡ x cos ⁡ 213 + cos ⁡ x sin ⁡ 213 sin ⁡ x cos ⁡ 213 − 1 2 ( cos ⁡ ( x − 213 ) − cos ⁡ ( x + 213 ) ) = cos ⁡ x sin ⁡ 213 + 1 2 ( cos ⁡ ( x − 213 ) + cos ⁡ ( x + 213 ) ) \begin{array}{l} \sin a\sin b = \frac{1}{2}\left( {\cos \left( {a - b} \right) - \cos \left( {a + b} \right)} \right)\\ \cos a\cos b = \frac{1}{2}\left( {\cos \left( {a - b} \right) + \cos \left( {a + b} \right)} \right)\\ \sin x\cos 213 - \sin x\sin 213 = \cos x\cos 213 + \cos x\sin 213\\ \sin x\cos 213 - \frac{1}{2}\left( {\cos \left( {x - 213} \right) - \cos \left( {x + 213} \right)} \right) = \cos x\sin 213 + \frac{1}{2}\left( {\cos \left( {x - 213} \right) + \cos \left( {x + 213} \right)} \right) \end{array}

With some rearrangement and another identity, we have:

sin ⁡ ( a − b ) = sin ⁡ a cos ⁡ b − sin ⁡ b cos ⁡ a sin ⁡ x cos ⁡ 213 − cos ⁡ x sin ⁡ 213 = cos ⁡ ( x − 213 ) sin ⁡ ( x − 213 ) = cos ⁡ ( x − 213 ) \begin{array}{l} \sin (a - b) = \sin a\cos b - \sin b\cos a\\ \sin x\cos 213 - \cos x\sin 213 = \cos \left( {x - 213} \right)\\ \sin \left( {x - 213} \right) = \cos \left( {x - 213} \right) \end{array}

Which means that :

sin ⁡ ( x − 213 ) = cos ⁡ ( x − 213 ) tan ⁡ ( x − 213 ) = 1 x − 213 = 45 x = 258 \begin{array}{l} \sin \left( {x - 213} \right) = \cos \left( {x - 213} \right)\\ \tan \left( {x - 213} \right) = 1\\ x - 213 = 45\\ x = 258 \end{array}

The last thing to do is verify if 258 is the smallest angle for x. Tan(x) is periodic within pi so we have :

tan ⁡ ( 258 ) ≡ tan ⁡ ( 78 ) \tan \left( {258} \right) \equiv \tan \left( {78} \right)

The final answer is x = 78 x = \boxed{78}

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Perhaps simpler, cos ⁡ 201 3 ∘ + sin ⁡ 201 3 ∘ cos ⁡ 201 3 ∘ − sin ⁡ 201 3 ∘ = cos ⁡ 3 3 ∘ + sin ⁡ 3 3 ∘ cos ⁡ 3 3 ∘ − sin ⁡ 3 3 ∘ = cos ⁡ 3 3 ∘ cos ⁡ 3 3 ∘ + sin ⁡ 3 3 ∘ cos ⁡ 3 3 ∘ cos ⁡ 3 3 ∘ cos ⁡ 3 3 ∘ − sin ⁡ 3 3 ∘ cos ⁡ 3 3 ∘ = 1 + tan ⁡ 3 3 ∘ 1 − tan ⁡ 3 3 ∘ = tan ⁡ 4 5 ∘ + tan ⁡ 3 3 ∘ 1 − tan ⁡ 4 5 ∘ tan ⁡ 3 3 ∘ = tan ⁡ ( 45 + 33 ) ∘ = tan ⁡ 7 8 ∘ \begin{aligned} \dfrac{\cos 2013^\circ+\sin 2013^\circ}{\cos 2013^\circ-\sin 2013^\circ}&=\dfrac{\cos 33^\circ+\sin 33^\circ}{\cos 33^\circ-\sin 33^\circ} \\ &=\dfrac{\dfrac{\cos 33^\circ}{\cos 33^\circ}+\dfrac{\sin 33^\circ}{\cos 33^\circ}}{\dfrac{\cos 33^\circ}{\cos 33^\circ}-\dfrac{\sin 33^\circ}{\cos 33^\circ}} \\ &=\dfrac{1+\tan 33^\circ}{1-\tan 33^\circ} \\ &=\dfrac{\tan 45^\circ+\tan 33^\circ}{1-\tan 45^\circ\tan 33^\circ} \\ &=\tan(45+33)^\circ \\ &=\tan 78^\circ \end{aligned} where the third-to-last step follows because tan ⁡ 4 5 ∘ = 1 \tan 45^\circ=1 , and the second-to-last step follows because tan ⁡ ( a + b ) = tan ⁡ a + tan ⁡ b 1 − tan ⁡ a tan ⁡ b \tan(a+b)=\dfrac{\tan a+\tan b}{1-\tan a\tan b} Hence, since the domain of arctan ⁡ \arctan is ( − 9 0 ∘ , 9 0 ∘ ) (-90^\circ,90^\circ) , 78 \boxed{78} is the smallest positive integer.

Daniel Chiu - 7 years, 7 months ago

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Nice one!

Isabelle Yeong - 7 years, 7 months ago

Nitpicking, but you're not simply concerned about the restricted domain, since we want the smallest positive integer. Instead, you want the answer to be in ( 1 , 180 (1, 180 .

Calvin Lin Staff - 7 years, 7 months ago

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Yeah, I was trying to think of a good way to say it. What I meant was:

tan ⁡ \tan is injective on the interval ( − 9 0 ∘ , 9 0 ∘ ) (-90^\circ,90^\circ) , and so no positive value below 7 8 ∘ 78^\circ has the same tangent as 7 8 ∘ 78^\circ , so the answer is 78 78 .

Daniel Chiu - 7 years, 7 months ago

cos ⁡ 201 3 ∘ + sin ⁡ 201 3 ∘ cos ⁡ 201 3 ∘ − sin ⁡ 201 3 ∘ = cos ⁡ 21 3 ∘ + sin ⁡ 21 3 ∘ cos ⁡ 21 3 ∘ − sin ⁡ 21 3 ∘ \dfrac{\cos 2013^{\circ}+\sin 2013^{\circ}}{\cos 2013^{\circ}-\sin 2013^{\circ}}=\dfrac{\cos 213^{\circ}+\sin 213^{\circ}}{\cos 213^{\circ}-\sin 213^{\circ}}

= cos ⁡ 3 3 ∘ + sin ⁡ 3 3 ∘ sin ⁡ 3 3 ∘ − cos ⁡ 3 3 ∘ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\dfrac{\cos 33^{\circ}+\sin 33^{\circ}}{\sin 33^{\circ}-\cos 33^{\circ}}

= ( cos ⁡ 3 3 ∘ + sin ⁡ 3 3 ∘ sin ⁡ 3 3 ∘ − cos ⁡ 3 3 ∘ ) ( cos ⁡ 3 3 ∘ + sin ⁡ 3 3 ∘ sin ⁡ 3 3 ∘ + cos ⁡ 3 3 ∘ ) \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\left( \dfrac{\cos 33^{\circ}+\sin 33^{\circ}}{\sin 33^{\circ}-\cos 33^{\circ}} \right) \left( \dfrac{\cos 33^{\circ}+\sin 33^{\circ}}{\sin 33^{\circ}+\cos 33^{\circ}} \right)

= 1 + 2 sin ⁡ 3 3 ∘ cos ⁡ 3 3 ∘ sin ⁡ 2 3 3 ∘ − cos ⁡ 2 3 3 ∘ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\dfrac{1+2 \sin 33^{\circ}\cos 33^{\circ}}{\sin^2 33^{\circ}-\cos^2 33^{\circ}}

= sin ⁡ 9 0 ∘ + sin ⁡ 6 6 ∘ − cos ⁡ 6 6 ∘ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\dfrac{\sin 90^{\circ}+\sin 66^{\circ}}{-\cos 66^{\circ}}

= 2 sin ⁡ 7 8 ∘ cos ⁡ 1 2 ∘ cos ⁡ 24 6 ∘ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\dfrac{2\sin 78^{\circ}\cos 12^{\circ}}{\cos 246^{\circ}}

= sin ⁡ 7 8 ∘ ( cos ⁡ 24 6 ∘ 2 cos ⁡ 1 2 ∘ ) \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\dfrac{\sin 78^{\circ}}{\left( \dfrac{\cos 246^{\circ}}{2\cos 12^{\circ}} \right)}

Now if we let ( cos ⁡ 24 6 ∘ 2 cos ⁡ 1 2 ∘ ) = cos ⁡ y \left( \dfrac{\cos 246^{\circ}}{2\cos 12^{\circ}} \right)=\cos y , we see that

cos ⁡ 24 6 ∘ = 2 cos ⁡ 1 2 ∘ cos ⁡ y \cos 246^{\circ}=2\cos 12^{\circ}\cos y

cos ⁡ 9 0 ∘ + cos ⁡ 24 6 ∘ = cos ⁡ ( 12 + y ) ∘ + c o s ( 12 − y ) ∘ \cos 90^{\circ}+\cos 246^{\circ}=\cos (12+y)^{\circ}+cos (12-y)^{\circ}

This gives

246 = 12 + y 246=12+y and 90 = 12 − y 90=12-y

Solving them for y we get y = 7 8 ∘ y=78^{\circ}

and hence

cos ⁡ 201 3 ∘ + sin ⁡ 201 3 ∘ cos ⁡ 201 3 ∘ − sin ⁡ 201 3 ∘ = sin ⁡ 7 8 ∘ ( cos ⁡ 24 6 ∘ 2 cos ⁡ 1 2 ∘ ) \dfrac{\cos 2013^{\circ}+\sin 2013^{\circ}}{\cos 2013^{\circ}-\sin 2013^{\circ}}=\dfrac{\sin 78^{\circ}}{\left( \dfrac{\cos 246^{\circ}}{2\cos 12^{\circ}} \right)}

= sin ⁡ 7 8 ∘ cos ⁡ 7 8 ∘ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\dfrac{\sin 78^{\circ}}{\cos 78^{\circ}}

= sin ⁡ 7 8 ∘ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\sin 78^{\circ}

Therefore, x = 7 8 ∘ x=78^{\circ}

Isabelle Yeong - 7 years, 7 months ago
Tunk-Fey Ariawan
Feb 3, 2014

The expression above can be rewritten as tan ⁡ x = 1 + sin ⁡ 2013 cos ⁡ 2013 1 − 1 ⋅ sin ⁡ 2013 cos ⁡ 2013 = tan ⁡ 4 5 ∘ + tan ⁡ 201 3 ∘ 1 − tan ⁡ 4 5 ∘ ⋅ tan ⁡ 201 3 ∘ = tan ⁡ ( 4 5 ∘ + 201 3 ∘ ) = tan ⁡ 205 8 ∘ . \begin{aligned} \tan x &=\frac{1+\frac{\sin 2013}{\cos 2013}}{1-1\cdot\frac{\sin 2013}{\cos 2013}}\\ &=\frac{\tan 45^\circ+\tan 2013^\circ}{1-\tan 45^\circ\cdot\tan 2013^\circ}\\ &=\tan (45^\circ + 2013^\circ)\\ &=\tan 2058^\circ. \end{aligned} Since tangent has a period of 18 0 ∘ 180^\circ , then 18 0 ∘ n + x = 205 8 ∘ 180^\circ n + x = 2058^\circ for n \;n integer. Therefore, x = 7 8 ∘ x=\boxed{78^\circ} . # Q . E . D . # \text{\#}\;\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}

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nice! :)

Omkar Sakhawalkar - 7 years, 3 months ago
Bhargav Das
Nov 12, 2013

We are given: tan ⁡ x ∘ = c o s 201 3 ∘ + sin ⁡ 201 3 ∘ c o s 201 3 ∘ − sin ⁡ 201 3 ∘ \tan x^\circ= \frac{cos 2013^\circ+\sin 2013^\circ}{cos 2013^\circ-\sin 2013^\circ} .

We follow the following steps:

STEP 1 : We divide both numerator and denominator by \(\cos 2013^\circ in Right Hand Side.

STEP **2**: We get,

\( \frac{1+\tan 2013^\circ}{1-\tan 2013^\circ}\)

= ( t a n 4 5 ∘ + tan ⁡ 201 3 ∘ 1 − tan ⁡ 201 3 ∘ × tan ⁡ 4 5 ∘ = (\frac{tan 45^\circ+\tan 2013^\circ}{1-\tan 2013^\circ \times \ \tan 45^\circ }

= tan ⁡ ( 2013 + 45 ) ∘ = tan ⁡ 205 8 ∘ = tan ⁡ ( 360 × 5 + 258 ) ∘ = tan ⁡ 25 8 ∘ = tan ⁡ ( 180 + 78 ) ∘ = tan ⁡ 7 8 ∘ = \tan (2013+45)^\circ=\tan 2058^\circ=\tan (360 \times 5+258)^\circ=\tan 258^\circ=\tan (180+78)^\circ=\tan 78^\circ

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The latex errors are due to carelessly pressing the Continue button.Sorry, for that.Can't edit:(

Bhargav Das - 7 years, 7 months ago

We want to rewrite the numerator cos ⁡ 201 3 o + sin ⁡ 201 3 o \cos 2013^{o} + \sin 2013^{o} in the form of R sin ⁡ ( θ + α ) R \sin (\theta + \alpha) . As R sin ⁡ ( θ + α ) = R ( sin ⁡ θ cos ⁡ α + cos ⁡ θ sin ⁡ α ) = cos ⁡ 201 3 o + sin ⁡ 201 3 o R \sin (\theta + \alpha) = R(\sin \theta \cos \alpha + \cos \theta \sin \alpha) = \cos 2013^{o} + \sin 2013^{o} , where θ = 201 3 o \theta = 2013^{o} , we want to have cos ⁡ α = sin ⁡ α \cos \alpha = \sin \alpha , or tan ⁡ α = 1 \tan \alpha = 1 . Here, α = 4 5 o \alpha = 45^{o} works. Note that we can use other valid values of α \alpha . Now, we want to have R cos ⁡ α = R sin ⁡ α = 1 R \cos \alpha = R \sin \alpha = 1 , thus R = 2 R = \sqrt{2} . Now, cos ⁡ 201 3 o + sin ⁡ 201 3 o = 2 ( sin ⁡ ( 2013 + 45 ) o ) = 2 ( sin ⁡ 205 8 o ) \cos 2013^{o} + \sin 2013^{o} = \sqrt{2}(\sin (2013+45)^{o}) = \sqrt{2}(\sin 2058^{o}) .

Similarly, we want to rewrite the denominator cos ⁡ 201 3 o − sin ⁡ 201 3 o \cos 2013^{o} - \sin 2013^{o} in the form of R cos ⁡ ( θ + α ) R \cos (\theta + \alpha) . By using a similar method as above, c o s 201 3 o − sin ⁡ 201 3 o = 2 ( sin ⁡ 205 8 o ) cos 2013^{o} - \sin 2013^{o} = \sqrt{2}(\sin 2058^{o}) .

Thus, tan ⁡ x o = sin ⁡ 205 8 o cos ⁡ 205 8 o = tan ⁡ 205 8 o = t a n 7 8 o \tan x^{o} = \frac{\sin 2058^{o}}{\cos 2058^{o}} = \tan 2058^{o} = tan 78^{o} . Hence, the smallest positive integer x x which satisfies the equation above is 78 \boxed{78} .

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Correction: cos ⁡ 201 3 o − sin ⁡ 201 3 o = 2 ( cos ⁡ 205 8 o ) \cos 2013^{o} - \sin 2013^{o} = \sqrt{2}(\cos 2058^{o}) , not 2 ( sin ⁡ 205 8 o ) \sqrt{2}(\sin 2058^{o}) .

Shendy Marcello Yuniar - 7 years, 5 months ago

great!!

andre yudhistika - 7 years, 5 months ago

2013 = 5 360 + 213 = 5 360 + 180 + 33 implies tan 2013 = tan 33 . So tan x = (1 + tan 33) /(1 - tan 33) = (tan45 + tan 33) /(1 - tan45 tan 33) = tan (45 + 33) = tan 78,
Hence x = 78 deg

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