Cosine of 90 degrees is zero.
Cos(90 - x) = sin(x) so cos2(90 - x) = sin2(x)
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°.
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.
cos(α) = sin(90° - α) → cos(16° + θ) = sin(90° - (16° + θ)) = sin(74° - θ) → sin(36° + θ) = cos(16° + θ) → sin((36° + θ) = sin(74° - θ) → 36° + θ = 74° - θ → 2θ = 38° → θ = 19° → θ = 19 °+ 180°n for n= 0, 1, 2, ...
No. Cos squared x is not the same as cos x squared. Cos squared x means cos (x) times cos (x) Cos x squared means cos (x squared)
sin θ = cos (90° - θ) cos θ = sin (90° - θ)
sin(30) = sin(90 - 60) = sin(90)*cos(60) - cos(90)*sin(60) = 1*cos(60) - 0*sin(60) = cos(60).
The expression (\cos^2(90^\circ - \theta)) can be simplified using the co-function identity, which states that (\cos(90^\circ - \theta) = \sin(\theta)). Therefore, (\cos^2(90^\circ - \theta) = \sin^2(\theta)). This means that (\cos^2(90^\circ - \theta)) is equal to the square of the sine of (\theta).
Cos(90 - x) = sin(x) so cos2(90 - x) = sin2(x)
To simplify the expression sin(30°) cos(90°) sin(90°) cos(30°), we first evaluate the trigonometric functions at the given angles. sin(30°) = 1/2, cos(90°) = 0, sin(90°) = 1, and cos(30°) = √3/2. Substituting these values into the expression, we get (1/2) * 0 * 1 * (√3/2) = 0. Therefore, the final result of sin(30°) cos(90°) sin(90°) cos(30°) is 0.
Cos 90=0 because cos@=base/hyp and at 90degree base becomes zero since zero divided by anything is zero. RPHK_Haider...
cos90=0
at -90 degrees the value of cos(x) is 0.
cosine (90- theta) = sine (theta)
The cofunction of the complement of cos 89° is sin 1°. This is because the complement of 89° is 1° (90° - 89° = 1°), and the cofunction identity states that (\cos(θ) = \sin(90° - θ)). Therefore, (\cos(89°) = \sin(1°)).
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°.
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.