Question

If sin A=4/5 and cos B=-12/13,
where A and B lie in first and third quadrant respectively, then
cos(A + B) =

​

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{sin\,A=\dfrac{4}{5}\;\;\&\;\;cos\,B=\dfrac{-12}{13}}$

$\textsf{A and B lies in the first and third quadrant respectively}$

$\underline{\textbf{To find:}}$

$\textsf{The value of}\;\mathsf{cos(A+B)}$

$\underline{\textbf{Solution:}}$

$\mathsf{\begin{array}{c|c}\mathsf{sin\,A=\dfrac{4}{5}}&\mathsf{cos\,B=\dfrac{-12}{13}}\\\\\mathsf{cos^2A=1-sin^2A}&\mathsf{sin^2B=1-cos^2B}\\\\\mathsf{cos^2A=1-\dfrac{16}{25}}&\mathsf{sin^2B=1-\dfrac{144}{169}}\\\\\mathsf{cos^2A=\dfrac{9}{25}}&\mathsf{sin^2B=\dfrac{25}{169}}\\\\\implies\mathsf{cosA=\pm\dfrac{3}{5}}&\implies\mathsf{sinB=\pm\dfrac{5}{13}}\\\\\end{array}}$

$\textsf{Since A lies in first quadrant, cosA is positive}$

$\implies\mathsf{cos\,A=\dfrac{3}{5}}$

$\textsf{Since B lies in third quadrant, sinB is negative}$

$\implies\mathsf{sin\,B=\dfrac{-5}{13}}$

$\mathsf{Now,}$

$\bf\;cos(A+B)=cos\,A\;cosB-sin\,A\;sinB$

$\mathsf{cos(A+B)=\dfrac{3}{5}\left(\dfrac{-12}{13}\right)-\dfrac{4}{5}\left(\dfrac{-5}{13}\right)}$

$\mathsf{cos(A+B)=\dfrac{-36}{65}+\dfrac{20}{65}}$

$\mathsf{cos(A+B)=\dfrac{-36+20}{65}}$

$\implies\boxed{\bf\,cos(A+B)=\dfrac{-16}{65}}$

Answer 2

Given:

sin A=4/5

cos B=-12/13,

A and B lie in first and third quadrant respectively

Solution:

Find the value of cos(A + B)

Solution:

we know that cos ( A+B) = cos A.cosB - SinA.SinB

now we have to find SinB and cos A

Cos A = √(1 - sin²A)

          =√(1 - (4/5)²)

          =√(25-16)/25

          =√9/25

 Cos A =3/5

as A is in first quadratnt so Cos A will be possitive.

Sin B = √(1 - cos²B)

         =√(1 - (-12/13)²)

         =√(13²-12²)/13²

         =√25/169

         =5/13

As B is in third quadrant therefore SinB will be negative

therefore , Sin B = -5/13

cos ( A+B) = cos A.cosB - SinA.SinB

                 =(3/5)(-12/13) - (4/5)(-5/13)

                 =( -36/65) + (20/65)

                  =-16/65

Hence the value of Cos ( A+B) is -16/65