Ratio Of Different Trigonometric Functions!

Geometry Level 3

If cos ⁡ 3 A cos ⁡ A \dfrac {\cos {3A}}{\cos {A}} = 1 2 = \dfrac {1}{2} , then the value of sin ⁡ 3 A sin ⁡ A \dfrac {\sin {3A}}{\sin {A}} can be expressed as a b \dfrac {a}{b} , where a a and b b are coprime positive integers , find a + b a+b .


The answer is 7.

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Rishabh Tiwari by Rishabh Tiwari , 20, India

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5 solutions

Sam Bealing
Jun 19, 2016

Relevant wiki: Triple Angle Identities

Using De Moivre's theorem we get:

cos ⁡ 3 A + i sin ⁡ 3 A = ( cos ⁡ A + i sin ⁡ A ) 3 = cos ⁡ 3 A + 3 i sin ⁡ A cos ⁡ 2 A − 3 sin ⁡ 2 A cos ⁡ A − i sin ⁡ 3 A \cos{3A}+i \; \sin{3A}=(\cos{A}+i \; \sin{A})^3=\cos^3{A}+3i \; \sin{A} \cos^2{A}-3 \sin^2{A} \cos{A}-i \; \sin^3{A}

cos ⁡ 3 A = cos ⁡ 3 A − 3 sin ⁡ 2 A cos ⁡ A sin ⁡ 3 A = 3 sin ⁡ A cos ⁡ 2 A − sin ⁡ 3 A \cos{3A}=\cos^3{A}-3\sin^2{A} \cos{A} \\ \sin{3A}=3\sin{A} \cos^2{A}-\sin^3{A}

cos ⁡ 3 A cos ⁡ A = cos ⁡ 2 A − 3 sin ⁡ 2 A = cos ⁡ 2 A − 3 ( 1 − cos ⁡ 2 A ) = 4 cos ⁡ 2 A − 3 \dfrac{\cos{3A}}{\cos{A}}=\cos^2{A}-3\sin^2{A}=\cos^2{A}-3 \left (1-\cos^2{A} \right )=4 \cos^2{A}-3

cos ⁡ 3 A cos ⁡ A = 1 2 ⟹ 4 cos ⁡ 2 A − 3 = 1 2 ⟹ cos ⁡ 2 A = 7 8 \dfrac{\cos{3A}}{\cos{A}}=\dfrac{1}{2} \implies 4 \cos^2{A}-3=\dfrac{1}{2} \implies \cos^2{A}=\dfrac{7}{8}

sin ⁡ 2 A = 1 − cos ⁡ 2 A = 1 − 7 8 = 1 8 \sin^2{A}=1-\cos^2{A}=1-\dfrac{7}{8}=\dfrac{1}{8}

sin ⁡ 3 A sin ⁡ A = 3 cos ⁡ 2 A − sin ⁡ 2 A = 21 8 − 1 8 = 20 8 = 5 2 \dfrac{\sin{3A}}{\sin{A}}= 3 \cos^2{A}-\sin^2{A}=\dfrac{21}{8}- \dfrac{1}{8}= \dfrac{20}{8}= \dfrac{5}{2}

a = 5 , b = 2 a=5,b=2

a + b = 7 \color{#20A900}{\boxed{\boxed{a+b=7}}}

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Moderator note:

Very clearly presented.

Nice solution , +1 :-)

Rishabh Tiwari - 4 years, 12 months ago

I tried to make 1/2= cos(pi/3), then inverse cousins whole thing. Obviously didn't work, can anyone help explain why.

Akira Kato - 4 years, 11 months ago

Sam , try out this problem . Its nice .

Rishabh Tiwari - 4 years, 12 months ago

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Good problem and good solution!

Sam Bealing - 4 years, 12 months ago

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Thank you :-)

Rishabh Tiwari - 4 years, 12 months ago
Ahmad Saad
Jun 19, 2016

==> a + b = 7

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C o s 3 A C o s A = C o s A ( 4 C o s 2 A − 3 ) C o s A = 4 ( 1 − S i n 2 A ) − 3 = ( − 4 S i n 2 A + 3 ) − 2 = S i n A ( − 4 S i n 2 A + 3 ) S i n A − 2 = 1 2 . ⟹ S i n 3 A S i n A = 5 2 = a b . \dfrac {Cos {3A}}{Cos {A}}= \dfrac {CosA(4Cos^2A-3)}{Cos {A}}=4(1 - Sin^2A) - 3\\ =( - 4Sin^2A + 3) - 2 =\dfrac{SinA( - 4Sin^2A + 3)}{SinA} - 2=\frac 1 2.\\ \implies\ \dfrac{Sin3A}{SinA}=\frac 5 2 =\frac a b.\\

So a + b = 5 + 2 = 7 a + b=5 + 2=\color{#D61F06}{7}

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@Niranjan Khanderia I have converted your comment into a solution.

I see that you did not answer this problem at all, which is why no solution box appears. In fact, you viewed the solution 15 seconds after viewing the problem. This contradicts what you said in "my answer was correct in the first try". Do you recall that you submitted the answer? If so, let me know what you did, so that I can track down the bug. Thanks!

Calvin Lin Staff - 4 years, 11 months ago

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Since I gave the right answer, I could see the solutions immediately. Thereafter I wrote the solution in Ahmad Saad comment.. That same day near about the same time , same thing happened to me for another problem.
If I remember correctly, Saad too faced the same problem or some thing like that since he too made some such comment as I remember correctly, though it might be for another problem I have mentioned. You may check records. there were more comments in this problem.
I clearly remember this problem I solved in one line with a few ==sgns and converting cos-square into sin-square in my rough calculation.
Thanks for your prompt action.


Niranjan Khanderia - 4 years, 11 months ago

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Thanks for the detailed information. I will investigate this issue further.

Calvin Lin Staff - 4 years, 11 months ago
Chew-Seong Cheong
Jul 12, 2016

cos ⁡ 3 A cos ⁡ A = 1 2 4 cos ⁡ 3 A − 3 cos ⁡ A cos ⁡ A = 1 2 4 cos ⁡ 2 A − 3 = 1 2 ⟹ cos ⁡ 2 A = 7 8 \begin{aligned} \frac {\cos 3A}{\cos A} & = \frac 12 \\ \frac {4\cos^3 A - 3\cos A}{\cos A} & = \frac 12 \\ 4\cos^2 A - 3 & = \frac 12 \\ \implies \cos^2 A & = \frac 78 \end{aligned}

Now, we have:

sin ⁡ 3 A sin ⁡ A = 3 sin ⁡ A − 4 sin ⁡ 3 A sin ⁡ A = 3 − 4 sin ⁡ 2 A = 3 − 4 ( 1 − cos ⁡ 2 A ) = 3 − 4 8 = 5 2 \begin{aligned} \frac {\sin 3A}{\sin A} & = \frac {3\sin A - 4\sin^3 A}{\sin A} \\ & = 3 - 4\sin^2 A \\ & = 3 - 4(1-\cos^2 A) \\ & = 3 - \frac 48 \\ & = \frac 52 \end{aligned}

⟹ a + b = 5 + 2 = 7 \implies a+b = 5+2 = \boxed{7}

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Not quite true. x 2 = 1 x^2 = 1 does not imply (only) that x = 1 x = 1 .

Calvin Lin Staff - 4 years, 11 months ago

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Don't get what you mean.

Chew-Seong Cheong - 4 years, 11 months ago

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cos ⁡ 2 A = 7 8 \cos^2 A = \frac{7}{8} does not imply that cos ⁡ A = 7 8 \cos A = \sqrt{ \frac{7}{8}} . Similarly, sin ⁡ 2 A = 1 8 \sin^2 A = \frac{1}{8} does not imply that sin ⁡ A = 1 8 \sin A = \sqrt{ \frac{1}{8} } .

Calvin Lin Staff - 4 years, 11 months ago

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@Calvin Lin – Thanks. I got it. I have changed the solution.

Chew-Seong Cheong - 4 years, 11 months ago

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@Chew-Seong Cheong – Looks good now. Thanks!

Calvin Lin Staff - 4 years, 11 months ago
Rocco Dalto
Oct 4, 2016

Using Euler's formula we obtain:

e i 3 θ = ( e i θ ) 3 ⟹ c o s ( 3 θ ) + i s i n ( 3 θ ) = ( c o s θ + i s i n θ ) 3 = {\bf e^{i3\theta} = (e^{i\theta})^3 \implies cos(3\theta) + isin(3\theta) = (cos\theta + isin\theta)^3 = }

c o s 3 θ + 3 c o s 2 θ s i n θ i − 3 c o s θ s i n 2 θ − s i n 3 θ i {\bf cos^3\theta + 3cos^2\theta sin\theta i - 3cos\theta sin^2\theta - sin^3\theta i }

⟹ c o s ( 3 θ ) = c o s 3 θ − 3 c o s θ s i n 2 θ = c o s θ ( c o s 2 θ − 3 s i n 2 θ ) = {\bf \implies cos(3\theta) = cos^3\theta - 3cos\theta sin^2\theta = cos\theta(cos^2\theta - 3sin^2\theta) = }

c o s θ ( 1 − s i n 2 θ − 3 s i n 2 ) = c o s ( 1 − 4 s i n 2 θ ) ⟹ {\bf cos\theta (1 - sin^2\theta - 3sin^2) = cos(1 - 4sin^2\theta) \implies }

c o s ( 3 θ ) c o s θ = 1 − 4 s i n 2 θ = 1 2 ⟹ s i n 2 θ = 1 2 ⟹ s i n θ = 1 2 2 {\bf \frac{cos(3\theta)}{cos\theta} = 1 - 4 sin^2\theta = \frac{1}{2} \implies sin^2\theta = \frac{1}{2} \implies sin\theta = \frac{1}{2\sqrt{2}} }

s i n 3 θ = s i n θ ( 3 c o s 2 θ − s i n 2 θ ) = s i n θ ( 3 ( 1 − s i n 2 θ ) − s i n 2 θ ) = s i n θ ( 3 − 4 s i n 2 θ ) ⟹ {\bf sin3\theta = sin\theta(3cos^2\theta - sin^2\theta) = sin\theta(3(1 - sin^2\theta) - sin^2\theta) = sin\theta(3 - 4 sin^2\theta) \implies }

s i n ( 3 θ ) s i n θ = 3 − 4 s i n 2 θ = 3 − 4 ( 1 8 ) = 3 − 1 2 = 5 2 {\bf \frac{sin(3\theta)}{sin\theta} = 3 - 4sin^2\theta = 3 - 4(\frac{1}{8}) = 3 - \frac{1}{2} = \boxed{\frac{5}{2}} }

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