Question

solve this sum.................​

Answer 1

Correct Question

if $Cos \alpha +Cos \beta$= $\frac{5}{4}$ and $sin \alpha +Sin \beta$=$\frac{1}{2}$

so find $Cos(\alpha-\beta)$

Given

$sin \alpha +Sin \beta$=$\frac{1}{2}$

$Cos \alpha +Cos \beta$=$\frac{5}{4}$

To Find

$Cos(\alpha-\beta)$

Solution

$sin \alpha +Sin \beta$=$\frac{1}{2}$

Both side squaring

=>($sin \alpha +Sin \beta$)²=($\frac{1}{2}$)²

$\implies{sin^2 \alpha +Sin^2 \beta++2sin \alpha \times sin\beta}$=$\frac{1}{4}$----(1)

Squaring again to the both side

$Cos \alpha +Cos \beta$=$\frac{5}{4}$

=>($Cos \alpha +Cos \beta$)²=($\frac{5}{4}$)²

$\implies{Cos^2 \alpha +Cos^2 \beta +2 cos \alpha \times cos\beta}$=$\frac{25}{16}$

Adding equation (1) and (2)

$\implies{sin^2 \alpha +Sin^2 \beta++2sin \alpha \times sin\beta}$ +${Cos^2 \alpha +Cos^2 \beta +2 cos \alpha \times cos\beta}$=$\frac{25}{16}$+$\frac{1}{4}$

$\implies{1+1+2sin \alpha \times sin\beta+2 cos \alpha \times cos\beta}$=$\frac{25+4}{16}$

$\implies 1+1+2sin \alpha \times sin\beta++2 cos \alpha \times cos\beta$=$\frac{29}{16}$

$\implies 2sin \alpha \times sin\beta+2 cos \alpha \times cos\beta$=$\frac{29}{16}$-2

$\implies{2(sin \alpha \times sin\beta+ cos \alpha \times cos\beta)}$=$\frac{29-32}{16}$

$\implies{sin \alpha \times sin\beta+cos \alpha \times cos\beta}$=$\frac{-3}{32}$

we know that

Cos(A-B)=cosA.CosB+SinA.SinB

Therefore

$Cos(\alpha-\beta)$=$\frac{-3}{32}$

Answer 2

Answer:

Cos(A-B)=cosA.CosB+SinA.SinB

Therefore

Cos(\alpha-\beta)Cos(α−β) =\frac{-3}{32} 32−3