If cosα + cosβ = (3)/(2)andsinα + sinβ = (1)/(2) and θ is the arithmetic mean of α and β, then 2θ + 2θ is equal to:

Correct answer: B. (7)/(5) Given cosα + cosβ = (3)/(2) sinα + sinβ = (1)/(2) ⇒ 2 ( (α + β)/(2) ) ( (α - β)/(2) ) = (3)/(2) ......(i) and 2 ( (α + β)/(2) ) ( (α - β)/(2) )= (1)/(2) ....(ii) Dividing (ii) and (i) (((α + β))/(2)) ∴ tan θ = (1)/(3) 2θ + 2θ = 2tan θ 1 + ^ 2 θ + 1 - ^ 2 θ 1 + ^ 2 θ = (2((1)/(3)))/(1 + (1)/(9)) + (1 - (1)/(9))/(1 + (1)/(9)) = ((2)/(3) + (8)/(9))/((10)/(9)) = (14)/(10) = (7)/(5)

JEE Main 201511 AprilOnline

Exam: JEE Main
Year: 2015
Date: 11 April
Mode: Online
Subject: Mathematics
Chapter: Trigonometric Equations

JEE Main 2015 · 11 April · Online · Mathematics · Trigonometric Equations

If cosα + cosβ = (3)/(2)andsinα + sinβ = (1)/(2) and θ is the...

cos\alpha + cos\beta = \dfrac{3}{2}\text{and}sin\alpha + sin\beta = \dfrac{1}{2}

and

\theta

is the arithmetic mean of

\alpha \text{ and } \beta,

then

\sin 2\theta + \cos 2\theta

is equal to:

Options

A $\dfrac{3}{5}$
B $\dfrac{7}{5}$
C $\dfrac{4}{5}$
D $\dfrac{8}{5}$

Correct answer

B. $\dfrac{7}{5}$

Step-by-step solution

Given

cos\alpha + cos\beta = \dfrac{3}{2}

sin\alpha + sin\beta = \dfrac{1}{2}

⇒ 2\cos ( \dfrac{\alpha + \beta}{2} ) \cos ( \dfrac{\alpha - \beta}{2} ) = \dfrac{3}{2}

......(i) and

2\sin ( \dfrac{\alpha + \beta}{2} )\cos ( \dfrac{\alpha - \beta}{2} )= \dfrac{1}{2}

....(ii)

Dividing (ii) and (i)

\tan \left(\dfrac{\left(\alpha + \beta\right)}{2}\right)

∴ tan \theta = \dfrac{1}{3}

\sin 2\theta + \cos 2\theta =

\dfrac{2tan \theta}{1 + \tan^{2} \theta} + \dfrac{1 - \tan^{2} \theta}{1 + \tan^{2} \theta} = \dfrac{2\left(\dfrac{1}{3}\right)}{1 + \dfrac{1}{9}} + \dfrac{1 - \dfrac{1}{9}}{1 + \dfrac{1}{9}}

= \dfrac{\dfrac{2}{3} + \dfrac{8}{9}}{\dfrac{10}{9}} = \dfrac{14}{10} = \dfrac{7}{5}

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