Question

cos 34 degrees + cos 64 degrees- cos 4 degrees ​

Answer 1

Answer:

answer is zero if it is subtracted nd 1 if multiplied

Answer 2

Answer:

$\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{Answer}}}}}}$

Prove that, \sin 34+\cos 64-\cos 4=0sin34+cos64−cos4=0

L.H.S. =\sin 34+\cos 64-\cos 4=sin34+cos64−cos4

=\sin 34+\cos (34+30)-\cos (34-30)=sin34+cos(34+30)−cos(34−30)

=\sin 34+\cos 34\cos 30-\sin 34\sin 30-(\cos 34\cos 30+\sin 34\sin 30)=sin34+cos34cos30−sin34sin30−(cos34cos30+sin34sin30)

Using trigonometric identity,

\cos (A+B)=\cos A\cos B-\sin A\sin Bcos(A+B)=cosAcosB−sinAsinB and

\cos (A-B)=\cos A\cos B+\sin A\sin Bcos(A−B)=cosAcosB+sinAsinB

=\sin 34+\cos 34\cos 30-\sin 34\sin 30-\cos 34\cos 30-\sin 34\sin 30=sin34+cos34cos30−sin34sin30−cos34cos30−sin34sin30

=\sin 34-\sin 34\sin 30-\sin 34\sin 30=sin34−sin34sin30−sin34sin30

=\sin 34-2\sin 34\sin 30=sin34−2sin34sin30

=\sin 34-2\sin 34(\dfrac{1}{2})=sin34−2sin34(21)

=\sin 34-\sin 34=sin34−sin34

= 0

= R.H.S., proved

Hence, \sin 34+\cos 64-\cos 4=0sin34+cos64−cos4=0 , proved.