answersLogoWhite

0

What is cos 90?

Updated: 4/28/2022
User Avatar

Wiki User

15y ago

Best Answer

Cosine of 90 degrees is zero.

User Avatar

Wiki User

15y ago
This answer is:
User Avatar

Add your answer:

Earn +20 pts
Q: What is cos 90?
Write your answer...
Submit
Still have questions?
magnify glass
imp
Related questions

What is the relationship between the cos and sin of the non 90 degree angles in a right angle triangle?

sin θ = cos (90° - θ) cos θ = sin (90° - θ)


Why is sine 30 the same as cosine of 60?

sin(30) = sin(90 - 60) = sin(90)*cos(60) - cos(90)*sin(60) = 1*cos(60) - 0*sin(60) = cos(60).


Cos square 90-x equals?

Cos(90 - x) = sin(x) so cos2(90 - x) = sin2(x)


How cos 90 is 0?

Cos 90=0 because cos@=base/hyp and at 90degree base becomes zero since zero divided by anything is zero. RPHK_Haider...


What is the value of cos -90?

cos90=0


What is cos 90 minus theta?

cosine (90- theta) = sine (theta)


What is the value of cosx at -90?

at -90 degrees the value of cos(x) is 0.


What trigonometric value is equal to cos 62?

The solution is found by applying the definition of complementary trig functions: Cos (&Theta) = sin (90°-&Theta) cos (62°) = sin (90°-62°) Therefore the solution is sin 28°.


Prove that sin 90 equals cos 50sin 40- cos 40 sin 50?

Sorry, but cos(50)sin(40) - cos(40)sin(50) is -0.1736, which is not even close to sin(90) which is 1.This does not work in radians, either. Please restate your question.


How tan9-tan27-tan63 tan81 equals 4?

tan(9) + tan(81) = sin(9)/cos(9) + sin(81)/cos(81)= {sin(9)*cos(81) + sin(81)*cos(9)} / {cos(9)*cos(81)} = 1/2*{sin(-72) + sin(90)} + 1/2*{sin(72) + sin(90)} / 1/2*{cos(-72) + cos(90)} = 1/2*{sin(-72) + 1 + sin(72) + 1} / 1/2*{cos(-72) + 0} = 2/cos(72) since sin(-72) = -sin(72), and cos(-72) = cos(72) . . . . . (A) Also tan(27) + tan(63) = sin(27)/cos(27) + sin(63)/cos(63) = {sin(27)*cos(63) + sin(63)*cos(27)} / {cos(27)*cos(63)} = 1/2*{sin(-36) + sin(90)} + 1/2*{sin(72) + sin(36)} / 1/2*{cos(-36) + cos(90)} = 1/2*{sin(-36) + 1 + sin(36) + 1} / 1/2*{cos(-36) + 0} = 2/cos(36) since sin(-36) = -sin(36), and cos(-36) = cos(36) . . . . . (B) Therefore, by (A) and (B), tan(9) - tan(27) - tan(63) + tan(81) = tan(9) + tan(81) - tan(27) - tan(63) = 2/cos(72) – 2/cos(36) = 2*{cos(36) – cos(72)} / {cos(72)*cos(36)} = 2*2*sin(54)*sin(18)/{cos(72)*cos(36)} . . . . . . . (C) But cos(72) = sin(90-72) = sin(18) so that sin(18)/cos(72) = 1 and cos(36) = sin(90-36) = sin(54) so that sin(54)/cos(36) = 1 and therefore from C, tan(9) – tan(27) – tan(63) + tan(81) = 2*2*1*1 = 4


What is tan of 90 degrees?

On the unit circle at 90 degrees the 90 degrees in radians is pi/2 and the coordinates for this are: (0,1). The tan function = sin/cos. In the coordinate system x is cos and y is sin. Therefore (0,1) ; cos=0, & sin=1 . Tan=sin/cos so tan of 90 degrees = 1/0. The answer of tan(90) = undefined. There can not be a 0 in the denominator, because you can't devide by something with no quantity. Something with no quantity is 0. Or, on a limits point of view, it would be infinity.


Can two non-zero vector got a zero resultant?

Yes.If the angle between them is 90 degrees. As we know that A.B=|A| |B| cos (phi). When phi=90 degree,cos 90=0. Hence A.B= |A| |B| *0 =0.