Question

Given: Sine (A) = four-fifths, StartFraction pi Over 2 EndFraction < A < Pi and Sine (B) = StartFraction negative 2 StartRoot 5 EndRoot Over 5 EndFraction, pi, Pi < B < StartFraction 3 pi Over 2 EndFraction

What is the value of cos(A – B)?

Negative StartFraction 2 StartRoot 5 EndRoot Over 25 EndFraction
Negative StartFraction StartRoot 5 EndRoot Over 5 EndFraction
StartFraction 2 StartRoot 5 EndRoot Over 5 EndFraction
StartFraction 11 StartRoot 5 EndRoot Over 25 EndFraction

Answer 1

Answer:

D

Step-by-step explanation:

Did test(edge 2020)

Answer 2

The value of $Cos(A - B)$ will be $(\frac{-\sqrt{5} }{5} )$.

What are trigonometric functions ?

Trigonometric functions are  the periodic functions which denote the relationship between angle and sides of a right-angled triangle.

We have,

$Sin(A) = \frac{4}{5}$ ,    $\frac{\pi }{2} < A < \pi$

$Sin(B) = \frac{-2\sqrt{5} }{5}$ ,   $\pi < B < \frac{3\pi }{2}$

So,

Using identity ;

$Cos(A - B)= CosA\ *\ Cos B + Sin A\ *\ Sin B$

So,

First find $SinB$ and $CosA$;

i.e.

$CosB=\frac{-\sqrt{5} }{5}=\frac{-1}{\sqrt{5} }$,   $\pi < B > \frac{3\pi }{2}$

$CosA=\frac{-3}{5}$,    $\frac{\pi }{2} < A > \pi$

Now,

Substituting values,

$Cos(A - B)= CosA\ *\ Cos B + Sin A\ *\ Sin B$

$Cos(A - B)= ((\frac{-3}{5})\ *\ \frac{-1}{\sqrt{5}} ) +((\frac{4}{5})\ *\ \frac{-2\sqrt{5} }{5} )$

Simplify,

$Cos(A - B)= (\frac{3 }{5\sqrt{5}} ) -(\frac{8 }{5\sqrt{5}})$

$Cos(A - B)= (\frac{-5 }{5\sqrt{5}} )$

$Cos(A - B)= (\frac{-1 }{\sqrt{5}} )$

Or we can write

$Cos(A - B)= (\frac{-\sqrt{5} }{5} )$

Hence, we can say that the value of $Cos(A - B)$ will be $(\frac{-\sqrt{5} }{5} )$ which is given in option (b).

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