If 0<x,y<π and cos⁡x+cos⁡y−cos⁡(x+y)=32 then sin⁡x+cos⁡y is equal to

2cos⁡x+y2cos⁡x−y2−2cos2⁡x+y2−1=32⇒ 2cos⁡x+y2cos⁡x−y2−2cos2⁡x+y2=12⇒4cos⁡x+y2cos⁡x−y2−4cos2⁡x+y2=1=cos2⁡x−y2+sin2⁡x−y2⇒4cos2⁡x+y2+cos2⁡x−y2−4cos⁡x+y2cos⁡x−y2+sin2⁡x−y2=0⇒cos⁡x−y2−2cos⁡x+y22+sin2⁡x−y2=0⇒ sin⁡x−y2=0⇒x=y⇒ cos⁡x−y2=2cos⁡x+y2 Now. put x=y ⇒ cos⁡x=12=cos⁡y∴ sin⁡x+cos⁡y=32+12=3+12

If 0<x,y<π and cos⁡x+cos⁡y−cos⁡(x+y)=32 then sin⁡x+cos⁡y is equal to

2cos⁡x+y2cos⁡x−y2−2cos2⁡x+y2−1=32⇒ 2cos⁡x+y2cos⁡x−y2−2cos2⁡x+y2=12⇒4cos⁡x+y2cos⁡x−y2−4cos2⁡x+y2=1=cos2⁡x−y2+sin2⁡x−y2⇒4cos2⁡x+y2+cos2⁡x−y2−4cos⁡x+y2cos⁡x−y2+sin2⁡x−y2=0⇒cos⁡x−y2−2cos⁡x+y22+sin2⁡x−y2=0⇒ sin⁡x−y2=0⇒x=y⇒ cos⁡x−y2=2cos⁡x+y2 Now. put x=y ⇒ cos⁡x=12=cos⁡y∴ sin⁡x+cos⁡y=32+12=3+12

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If $0 < x, y < π and \cos x + \cos y - \cos (x + y) = \frac{3}{2}$ then $\sin x + \cos y$ is equal to

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{1 - \sqrt{3}}{2}$

$\frac{1 + \sqrt{3}}{2}$

$\frac{1}{2}$

$\frac{\sqrt{3}}{2}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{aligned} 2 \cos (\frac{x + y}{2}) \cos (\frac{x - y}{2}) - [2 \cos^{2} (\frac{x + y}{2}) - 1] = \frac{3}{2} \\ \Rightarrow 2 \cos (\frac{x + y}{2}) \cos (\frac{x - y}{2}) - 2 \cos^{2} (\frac{x + y}{2}) = \frac{1}{2} \end{aligned}$

$\begin{array}{l} \Rightarrow 4 \cos (\frac{x + y}{2}) \cos (\frac{x - y}{2}) - 4 \cos^{2} (\frac{x + y}{2}) = 1 \\ = \cos^{2} (\frac{x - y}{2}) + \sin^{2} (\frac{x - y}{2}) \\ \Rightarrow 4 \cos^{2} (\frac{x + y}{2}) + \cos^{2} (\frac{x - y}{2}) - 4 \cos (\frac{x + y}{2}) \cos (\frac{x - y}{2}) + \sin^{2} (\frac{x - y}{2}) = 0 \end{array}$

$\begin{array}{l} \Rightarrow {(\cos (\frac{x - y}{2}) - 2 \cos (\frac{x + y}{2}))}^{2} + \sin^{2} (\frac{x - y}{2}) = 0 \\ \Rightarrow \sin (\frac{x - y}{2}) = 0 \Rightarrow x = y \\ \Rightarrow \cos (\frac{x - y}{2}) = 2 \cos (\frac{x + y}{2}) Now. put x=y \\ \Rightarrow \cos x = \frac{1}{2} = \cos y \\ ∴ \sin x + \cos y = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{\sqrt{3} + 1}{2} \end{array}$