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Geometry Level 3

If 5 tan ⁡ ( α ) tan ⁡ ( β ) = 3 5 \tan(\alpha)\tan(\beta)=3 , then find the value of cos ⁡ ( α + β ) cos ⁡ ( α − β ) \dfrac{\cos\left(\alpha+\beta\right)}{\cos\left(\alpha-\beta\right)}

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The answer is 0.25.

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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4 solutions

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Sandeep Bhardwaj
Dec 2, 2014

Given : 5. t a n α . t a n β = 3 5.tan\alpha.tan\beta=3

⟹ 5. s i n α . s i n β c o s α . c o s β = 3 \implies \dfrac{5.sin\alpha.sin\beta}{cos\alpha.cos\beta}=3

⟹ s i n α . s i n β c o s α . c o s β = 3 5 \implies \dfrac{sin\alpha.sin\beta}{cos\alpha.cos\beta}=\dfrac{3}{5}

Using componendo and dividendo,

c o s α . c o s β + s i n α . s i n β c o s α . c o s β − s i n α . s i n β = 5 + 3 5 − 3 = 2 \dfrac{cos\alpha.cos\beta+sin\alpha.sin\beta}{cos\alpha.cos\beta-sin\alpha.sin\beta}=\dfrac{5+3}{5-3}=2

⟹ c o s ( α − β ) c o s ( α + β ) = 4 \implies \dfrac{cos\left(\alpha-\beta\right)}{cos\left(\alpha+\beta\right)}=4

T h e r e f o r e , c o s ( α + β ) c o s ( α − β ) = 1 4 = 0.25 Therefore, \quad \dfrac{cos\left(\alpha+\beta\right)}{cos\left(\alpha-\beta\right)}=\dfrac{1}{4}=\boxed{0.25}

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What does Compendo and Divendo mean? Could someone please explain?

Ritu Roy - 6 years, 6 months ago

Componendo and Dividendo means : If a b = c d \dfrac{a}{b}=\dfrac{c}{d} , then we can say that a + b a − b = c + d c − d \dfrac{a+b}{a-b}=\dfrac{c+d}{c-d} .That means that we can apply the + , − +,- operators on both the sides with Numerators and Denominators. @Ritu Roy

Sandeep Bhardwaj - 6 years, 6 months ago

Here is a note that I wrote up in the past about how to use it.

Calvin Lin Staff - 6 years, 6 months ago

I went the other way round . I divided the numerator and denominator in cos(a-b)/cos(a+b) by cos(a)cos(b) and then substituted the value of tan(a)tan(b).

Keshav Tiwari - 6 years, 6 months ago

exactly the same!

Asim Das - 6 years, 5 months ago

My solution involves the Product-to-Sum formulas:

sin ⁡ ( u ) sin ⁡ ( v ) = 1 2 [ cos ⁡ ( u − v ) − cos ⁡ ( u + v ) ] \sin { \left( u \right) } \sin { \left( v \right) } =\frac { 1 }{ 2 } \left[ \cos { \left( u-v \right) } -\cos { \left( u+v \right) } \right]

cos ⁡ ( u ) cos ⁡ ( v ) = 1 2 [ cos ⁡ ( u − v ) + cos ⁡ ( u + v ) ] \cos { \left( u \right) } \cos { \left( v \right) } =\frac { 1 }{ 2 } \left[ \cos { \left( u-v \right) } +\cos { \left( u+v \right) } \right]

Given: 5 tan ⁡ ( α ) tan ⁡ ( β ) = 3 5\tan { \left( \alpha \right) } \tan { \left( \beta \right) } =3

⟹ 5 sin ⁡ α sin ⁡ β cos ⁡ α cos ⁡ β = 3 \Longrightarrow \quad 5\frac { \sin { \alpha } \sin { \beta } }{ \cos { \alpha } \cos { \beta } } \quad =\quad 3

⟹ 5 sin ⁡ α sin ⁡ β = 3 cos ⁡ α cos ⁡ β \Longrightarrow \quad 5\sin { \alpha } \sin { \beta } \quad =\quad 3\cos { \alpha } \cos { \beta }

Using said formulas, we get

5 × 1 2 [ cos ⁡ ( α − β ) − cos ⁡ ( α + β ) ] = 3 × 1 2 [ cos ⁡ ( α − β ) + cos ⁡ ( α + β ) ] ⟹ 5 cos ⁡ ( α − β ) − 5 cos ⁡ ( α + β ) = 3 cos ⁡ ( α − β ) + 3 cos ⁡ ( α + β ) ⟹ 2 cos ⁡ ( α − β ) = 8 cos ⁡ ( α + β ) ⟹ cos ⁡ ( α + β ) cos ⁡ ( α − β ) = 2 8 = 0.25 5\quad \times \quad \frac { 1 }{ 2 } \left[ \cos { \left( \alpha -\beta \right) } -\cos { \left( \alpha +\beta \right) } \right] \quad =\quad 3\quad \times \quad \frac { 1 }{ 2 } \left[ \cos { \left( \alpha -\beta \right) } +\cos { \left( \alpha +\beta \right) } \right] \\ \Longrightarrow \quad 5\cos { \left( \alpha -\beta \right) } \quad -\quad 5\cos { \left( \alpha +\beta \right) } \quad =\quad 3\cos { \left( \alpha -\beta \right) } \quad +\quad 3\cos { \left( \alpha +\beta \right) } \\ \Longrightarrow \quad 2\cos { \left( \alpha -\beta \right) } \quad =\quad 8\cos { \left( \alpha +\beta \right) } \\ \Longrightarrow \quad \frac { \cos { \left( \alpha +\beta \right) } }{ \cos { \left( \alpha -\beta \right) } } \quad =\quad \frac { 2 }{ 8 } \quad =\quad \boxed { 0.25 }

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Akhil Kumar
Dec 6, 2014

cos(a+b)/cos(a-b) = (cosa.cosb-sina.sinb)/(cosa.cosb+sina.sinb) Now dividing numerator and denominator by sina.sinb,we get (cota.cotb-1)/(cota.cotb+1) now putting cota.cotb=5/3 (Given) we get, (5/3-1)/(5/3-1)=1/4=0.25

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Well I solved it with some different way, I am considering alpha as A and beta as B as tanAtanB=3/5 ------- let Eq No 1, now opening cos(A+B)/cos(A-B)=cosAcosB-SinAsinB/cosAcosB+sinAsinB dividing numerator and denominator by cosAcosB. =1-tanAtanB/1+tanAtanB =1-3/5 / 1 + 3/5 =2/8=0.25

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Perfect . .......,.............

Wael Noaman - 6 years, 4 months ago

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