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The inverse (negatives) of sine, cosine, and tangent are used to calculate the angle theta (or whatever you choose to name it).
Initially it is taught that opposite over hypotenuse is equal to the sine of theta
sin(theta) = opposite/hypotenuse
So it can be said that
theta = sin-1 (opp/hyp)
This works the same way with cosine and tangent
In short the inverse is simply what you use when you move the sin, cos, or tan to the other side of the equation generally to find the angle
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How would you prove left cosA plus sinA right times left cos2A plus sin2A right equals cosA plus sin3A?
You need to make use of the formulae for sin(A+B) and cos(A+B), and that cos is an even function: sin(A+B) = cos A sin B + sin A cos B cos(A+B) = cos A cos B - sin A sin B cos even fn → cos(-x) = cos(x) To prove: (cos A + sin A)(cos 2A + sin 2A) = cos A + sin 3A The steps are to work with the left hand side, expand the brackets, collect [useful] terms together, apply A+B formula above (backwards) and apply even nature of cos function: (cos A + sin A)(cos 2A + sin 2A) = cos A cos 2A + cos A sin 2A + sin A cos 2A + sin A sin 2A = (cos A cos 2A + sin A sin 2A) + (cos A sin 2A + sin A cos 2A) = cos(A - 2A) + sin(A + 2A) = cos(-A) + sin 3A = cos A + sin 3A which is the right hand side as required.
Find y'' if y equals 6x sinx?
That means you must take the derivative of the derivative. In this case, you must use the product rule. y = 6x sin x y'= 6[x (sin x)' + (x)' sin x] = 6[x cos x + sin x] y'' = 6[x (cos x)' + (x)' cos x + cos x] = 6[x (-sin x) + cos x + cos x] = 6[-x sin x + 2 cos x]
Cos 195 degree use the trigonometry functions of quadrantal angles and special to find the exact value?
cos(195) = cos(180 + 15) = cos(180)*cos(15) - sin(180)*sin(15) = -1*cos(15) - 0*sim(15) = -cos(15) = -cos(60 - 45) = -[cos(60)*cos(45) + sin(60)*sin(45)] = -(1/2)*sqrt(2)/2 - sqrt(3)/2*sqrt(2)/2 = - 1/4*sqrt(2)*(1 + sqrt3) or -1/4*[sqrt(2) + sqrt(6)]
Given that x is acute and cos x 0.6 find tan x?
Oh, dude, you're hitting me with some math lingo there! So, if x is acute and cos x is 0.6, we can use the Pythagorean identity to find sin x, which is 0.8. Then, to find tan x, we just divide sin x by cos x, giving us 0.8 / 0.6, which simplifies to 4/3. So, tan x is 4/3. Math can be cool, but like, let's not get too serious about it, right?
Why is tan90 equals infinity?
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.
