The limit does not exist.
sin(x)-cos(x) = (1)sin(x)+(-1)cos(x) so the range is sqrt((1)^2+(-1)^2)=1 and the domain is R <><><><><> The domain of sin x - cos x is [-infinity, +infinity]. The range of sin x - cos x is [-1.414, +1.414].
If I understand your question correctly, such a sequence is an = x cos(πx). It has neither an upper nor lower bound. It's divergent, but its limit is neither infinity nor negative infinity.
tan x = sin x / cos x sin x = x - (x^3/3!) + (x^5/5!) - (x^7/7!) + (x^9/9!) ... to infinity cos x = 1 - (x^2/2!) + (x^4/4!) - (x^6/6!) + (x^8/8!) .. to infinity tan 165 degrees = -0.267949192
Cos times Cos
The range for y = 4 cos (2x) is [-4, +4].Not asked, but answered for completeness sake, the domain is [-infinity, +infinity].
The limit does not exist.
sin(x)-cos(x) = (1)sin(x)+(-1)cos(x) so the range is sqrt((1)^2+(-1)^2)=1 and the domain is R <><><><><> The domain of sin x - cos x is [-infinity, +infinity]. The range of sin x - cos x is [-1.414, +1.414].
If I understand your question correctly, such a sequence is an = x cos(πx). It has neither an upper nor lower bound. It's divergent, but its limit is neither infinity nor negative infinity.
The answer depends on the side from which x approaches 0. If from the negative side, then the limit is negative infinity whereas if from the positive side, it is positive infinity.
tan x = sin x / cos x sin x = x - (x^3/3!) + (x^5/5!) - (x^7/7!) + (x^9/9!) ... to infinity cos x = 1 - (x^2/2!) + (x^4/4!) - (x^6/6!) + (x^8/8!) .. to infinity tan 165 degrees = -0.267949192
Note: Assuming you are working with natural, integer, rational(fraction), or real numbers. It doesn't. Infinity is not a number, even though, due to us mathematicians being lazy, we denote something = infinity. But we NEVER write tan 90 = infinity. But rather lim_x->(90degree) tan x = infinity. Meaning as x gets closer to 90 degree (even though degree is a horrible measurement for angle, we will use it), the value of tan x gets large faster and unbounded. tan x? It doesn't exist. Why? Because tan x is defined as (sin x / cos x). When x = 90 degree, cos x = 0, while sin x is positive around x = 90 degree. sin x / cos x := sinx x 1 / cos x, x = 90 degree, we get 1 x 1 / 0. But the definition for inverses does not include 0, meaning 1 / 0 does NOT exist. so, sadly, tan 90degree doesn't exist. The best we can do is see what happens around x = 90degree for tan, as you go into Calculus, you will know the tool is called limits and derivatives. And you might also see the reason it is +infinity, but not -infinity. (tan x approaches -infinity as x approaches 180degree). WHat's more? You will learn a far better measurement for angle that you will stick with in Calculus.
For the product to be zero, any of the factors must be zero, so you solve, separately, the two equations: sin x = 0 and: cos x = 0 Like many trigonometric equations, this will have an infinity of solutions, since sine and cosine are periodic functions.
I think so cos There is this adorable pug that lives across the road called Ike and he is amazingly good with people aged 0-infinity! he is so cute too!
Cos times Cos
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)
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.