Sine and Cosine ratios

Geometry Level 2

sin ⁡ 2 ( 3 A ) sin ⁡ 2 ( A ) − cos ⁡ 2 ( 3 A ) cos ⁡ 2 ( A ) = 2 \large \dfrac{\sin^2 (3A)}{\sin^2 (A)}-\dfrac{\cos^2 (3A)}{\cos^2 (A)}=2

⟹ cos ⁡ ( 2 A ) = ? \implies \large \cos (2A)= \ ?

1 4 \frac{1}{4} 1 2 \frac{1}{2} 1 2 \frac{1}{\sqrt{2}} 1 3 \frac{1}{\sqrt{3}}

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Ikkyu San by Ikkyu San , 23, Thailand

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6 solutions

Hem Shailabh Sahu
Apr 20, 2015

L . H . S . = s i n 2 ( 3 A ) s i n 2 ( A ) − c o s 2 ( 3 A ) c o s 2 ( A ) L.H.S.=\frac{sin^2(3A)}{sin^2(A)}-\frac{cos^2(3A)}{cos^2(A)}

= s i n 2 ( 3 A ) . c o s 2 ( A ) − c o s 2 ( 3 A ) . s i n 2 ( A ) s i n 2 ( A ) . c o s 2 ( A ) =\frac{sin^2(3A).cos^2(A)-cos^2(3A).sin^2(A)}{sin^2(A).cos^2(A)}

= [ s i n ( 3 A ) . c o s ( A ) − c o s ( 3 A ) . s i n ( A ) ] . [ s i n ( 3 A ) . c o s ( A ) + c o s ( 3 A ) . s i n ( A ) ] s i n 2 ( A ) . c o s 2 ( A ) =\frac{[sin(3A).cos(A)-cos(3A).sin(A)].[sin(3A).cos(A)+cos(3A).sin(A)]}{sin^2(A).cos^2(A)}

Using the identities : ( i ) s i n ( A ± B ) = s i n A . c o s B ± c o s A . s i n B (i) sin(A \pm B)=sinA.cosB \pm cosA.sinB & ( i i ) s i n ( 2 θ ) = 2. s i n θ . c o s θ (ii) sin(2\theta)=2.sin\theta.cos\theta

= s i n ( 3 A − A ) . s i n ( 3 A + A ) s i n 2 ( A ) . c o s 2 ( A ) = s i n ( 2 A ) . s i n ( 4 A ) s i n 2 ( A ) . c o s 2 ( A ) =\frac{sin(3A-A).sin(3A+A)}{sin^2(A).cos^2(A)}=\frac{sin(2A).sin(4A)}{sin^2(A).cos^2(A)}

= s i n ( 2 A ) . ( 2. s i n ( 2 A ) . c o s ( 2 A ) ) s i n 2 ( A ) . c o s 2 ( A ) =\frac{sin(2A).(2.sin(2A).cos(2A))}{sin^2(A).cos^2(A)}

= 8. s i n 2 ( A ) . c o s 2 ( A ) . c o s ( 2 A ) s i n 2 ( A ) . c o s 2 ( A ) = 8. c o s ( 2 A ) =\frac{8.sin^2(A).cos^2(A).cos(2A)}{sin^2(A).cos^2(A)}=8.cos(2A)

R . H . S . = 2 R.H.S.=2

⇒ 8. c o s ( 2 A ) = 2 ⇒ c o s ( 2 A ) = 1 4 \Rightarrow 8.cos(2A)=2 \Rightarrow cos(2A)=\boxed{\frac{1}{4}}

:D

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Moderator note:

Great!

I just used sin^2(a)+cos^2(a)=1 then completely it became simple. Double angle formula was much more simpler

Aayush Patni - 6 years, 1 month ago

Yup I too did it the same way.Great....

Raushan Sharma - 6 years, 1 month ago

An easier method..... Sin2(3A)/sin2(A) - cos23A/cosA=2 LHS =(3sinA-4sin3A)2/sin2A - (4cos3A-3cosA)2/cos2A =sin2A(3-4sin2A)2/sin2A - cos2A(4cos2A-3)2/cos2A =(3-4sinA)2 – (4cos2A-3)2 =(3-4 sin sq.A + 4cos sq.A-3)(3-4sin sq.A-4cos sq.A + 3) =4(cos sq.A-sin sq.A)(6-4) 4(cos2A)(2)=2 Cos2A=1/4

Manisha Bhosley - 6 years, 1 month ago

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Triple Angle Formula... That was what Pi Han Goh's solution is based on too...

Hem Shailabh Sahu - 6 years, 1 month ago

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Ohh... I didn't see it.. :P I just solved my way and posted the soln. I didn't see Pi Han's soln.

Manisha Bhosley - 6 years, 1 month ago
Pi Han Goh
Apr 20, 2015

Recall the triple angle identities: sin ⁡ ( 3 A ) = − 4 sin ⁡ 3 A + 3 sin ⁡ A , cos ⁡ ( 3 A ) = 4 cos ⁡ 3 A − 3 cos ⁡ A \sin(3A) = -4\sin^3 A + 3\sin A, \cos(3A) = 4\cos^3 A - 3\cos A .

For simplicity sake, let s = sin ⁡ ( A ) , c = cos ⁡ ( A ) s = \sin(A), c = \cos(A) , so s 2 + c 2 = 1 s^2+c^2=1 and we want to find c 2 − s 2 c^2 - s^2 .

( − 4 s 2 + 3 ) 2 − ( 4 c 2 − 3 ) 2 = 2 16 ( s 4 − c 4 ) − 24 ( s 2 − c 2 ) = 2 16 ( s 2 − c 2 ) ( s 2 + c 2 ) − 24 ( s 2 − c 2 ) = 2 ( 16 − 24 ) ( s 2 − c 2 ) = 2 c 2 − s 2 = − 2 16 − 24 = 1 4 \begin{aligned} (-4s^2+ 3)^2 - (4c^2 - 3)^2 &=& 2 \\ 16(s^4-c^4) - 24(s^2-c^2) &=& 2 \\ 16(s^2-c^2)(s^2+c^2) - 24(s^2-c^2) &=& 2 \\ (16-24)(s^2-c^2) &=& 2 \\ c^2-s^2 &=& -\frac {2}{16-24} = \boxed { \frac 1 4 } \end{aligned}

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Moderator note:

Yes. Triple Angle Identities is a simple approach.

Chew-Seong Cheong
Apr 20, 2015

sin ⁡ 2 3 A sin ⁡ 2 A − cos ⁡ 2 3 A cos ⁡ 2 A = 2 ⇒ sin ⁡ 2 3 A cos ⁡ 2 A − cos ⁡ 2 3 A sin ⁡ 2 A sin ⁡ 2 A cos ⁡ 2 A = 2 ⇒ ( sin ⁡ 3 A cos ⁡ A − cos ⁡ 3 A sin ⁡ A ) ( sin ⁡ 3 A cos ⁡ A + cos ⁡ 3 A sin ⁡ A ) = 2 sin ⁡ 2 A cos ⁡ 2 A ⇒ sin ⁡ 2 A sin ⁡ 4 A = 1 2 sin ⁡ 2 2 A ⇒ 2 sin ⁡ 2 2 A cos ⁡ 2 A = 1 2 sin ⁡ 2 2 A ⇒ cos ⁡ 2 A = 1 4 \dfrac{\sin^2{3A}}{\sin^2{A}}-\dfrac{\cos^2{3A}}{\cos^2{A}} = 2 \quad \Rightarrow \dfrac {\sin^2{3A}\cos^2{A}-\cos^2{3A}\sin^2{A}}{\sin^2{A}\cos^2{A}} = 2 \\ \Rightarrow (\sin{3A}\cos{A}-\cos{3A}\sin{A})(\sin{3A}\cos{A}+\cos{3A}\sin{A}) = 2\sin^2{A}\cos^2{A}\\ \Rightarrow \sin{2A} \sin{4A} = \dfrac{1}{2}\sin^2{2A} \quad \Rightarrow 2\sin^2{2A} \cos{2A} = \dfrac{1}{2}\sin^2{2A} \\ \Rightarrow \cos{2A} = \boxed{\frac{1}{4}}

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Moderator note:

The use of difference of squares to compound angle formula nailed the problem. Well done.

I just loved your Solution!

Skanda Prasad - 3 years, 7 months ago
Rocco Dalto
Sep 30, 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) = }

1 2 ∗ c o s θ ( 1 + c o s ( 2 θ ) − 3 ( 1 − c o s ( 2 θ ) ) = {\bf \frac{1}{2} * cos\theta (1 + cos(2\theta) - 3(1 - cos(2\theta)) = }

c o s θ ( 2 c o s ( 2 θ ) − 1 ) {\bf cos\theta (2cos(2\theta) - 1) }

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

⟹ 2 = ( 2 c o s ( 2 A ) + 1 ) 2 − ( 2 c o s ( 2 A ) − 1 ) 2 = 8 c o s ( 2 A ) ⟹ {\bf \implies 2 = (2cos(2A) + 1)^2 - (2cos(2A) - 1)^2 = 8cos(2A) \implies }

c o s ( 2 A ) = 1 4 {\bf cos(2A) = \frac{1}{4} }

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Otto Bretscher
Apr 21, 2015

The triple angle and double angle identities give sin ⁡ 3 A = ( sin ⁡ A ) ( 3 − 4 sin ⁡ 2 A ) = ( sin ⁡ A ) ( 2 cos ⁡ 2 A + 1 ) \sin{3A}=(\sin{A)}(3-4\sin^2{A})=(\sin{A})(2\cos{2A}+1) and cos ⁡ 3 A = ( cos ⁡ A ) ( 4 cos ⁡ 2 A − 3 ) = ( cos ⁡ A ) ( 2 cos ⁡ 2 A − 1 ) \cos{3A}=(\cos{A})(4\cos^2{A}-3)=(\cos{A})(2\cos{2A}-1) .

Thus ( sin ⁡ 3 A sin ⁡ A ) 2 − ( cos ⁡ 3 A cos ⁡ A ) 2 = ( 2 cos ⁡ 2 A + 1 ) 2 − ( 2 cos ⁡ 2 A − 1 ) 2 (\frac{\sin{3A}}{\sin{A}})^2- (\frac{\cos{3A}}{\cos{A}})^2 =(2\cos{2A}+1)^2-(2\cos{2A}-1)^2 = 8 cos ⁡ 2 A = 2 =8\cos{2A}=2 and cos ⁡ 2 A = 1 4 \cos{2A}=\boxed{\frac{1}{4}}

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Moderator note:

The addition of double angle identities certainly did simplify the work. Great job!

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