f cos A =4/5 and cos B = 12/13, what is the value of cos (A – B) ? [Note: cos (A – B) = cos A cos?

Question

f cos A =4/5 and cos B = 12/13, what is the value of cos (A – B) ? [Note: cos (A – B) = cos A cos
B + sin A sin B.]

Answer 1

Your formula is correct.
sin(A)=sqrt(1-cos^2(A))
cos^2(A)=cos(A)cos(A)

Answer 2

Draw a rt angle triangle involving cos(a) Use Pythag th. then other side is 3. Therefore sin(a) = 3/5. Do the same with cos(b) then other side is sqrt(169-144)=5. Sin(b) = 5/13

cos(a-b) = cos(a)cos(b)+sin(a)sin(b)
cos(a-b)=(4/5)(3/5)+(3/5)(5/13)
cos(a-b)=(12/25)+(15/65) and hopefully you can finish it yourself. Watch out for any mistakes I may have made.

Answer 3

Given; cosA=4/5. therefore find
sinA=sqr of 1-cos^2A ;
= sqr of 1-{4/5}^2 ;
=sqrof 1-(16/25)=3/5;
-similarly, sinB=5/13
cos(A-B)= cosAcosB+sinAsinB
=(4/5)(12/13)+(3/5)(5/13)
={60+15}/13x5=75/65
=15/13
thankx for giving an oppturnity....

Answer 4

= 4/5 *12/13 + 3/5*5/13
=48/65+ 3/13
=48/65 + 15/65 = 63/65

Answer 5

cosacosb=4/5 x12/13=48/65
sina=3/5, sinb=5/13 use pythagoreaus theorem
sinasinb=3/13
cos(a-b)=48/65-3/13=48/65-15/65=33/65