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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)]
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.
Why we use sine angle in cross product while cos angle in dot product?
Because in dot product we take projection fashion and that is why we used cos and similar in cross product we used sin
How do you simplify tan cot equals 1?
It just simplifies down to 1=1. You have to use your trig identities... tan=sin/cos cot=cos/sin thus tan x cot= (sin/cos) (cos/sin) since sin is in the numerator for tan, when it is multiplied by cot (which has sin in the denominator) both of the signs cancel and both now have a value of 1. The same happens with cos. so you get 1 x 1=1 so there is your answer. just learn your trig identities and you will understand
How To Use SohCahToa?
Sin= Opposite/Hypotenuse Cos= Adjacent/Hypotenuse Tan= Opposite/Adjacent
How do you get angle size of a 60 feet multiply 90 feet?
5400
How do you solve trignometric identities?
tan(x) = sin(x)/cos(x) Therefore, all trigonometric ratios can be expressed in terms of sin and cos. So the identity can be rewritten in terms of sin and cos. Then there are only two "tools": sin^2(x) + cos^2(x) = 1 and sin(x) = cos(pi/2 - x) Suitable use of these will enable you to prove the identity.
Cos plus tansin equals sec?
Start on the left-hand side. cos(x) + tan(x)sin(x) Put tan(x) in terms of sin(x) and cos(x). cos(x) + [sin(x)/cos(x)]sin(x) Multiply. cos(x) + sin2(x)/cos(x) Make the denominators equal. cos2(x)/cos(x) + sin2(x)/cos(x) Add. [cos2(x) + sin2(x)]/cos(x) Use the Pythagorean Theorem to simplify. 1/cos(x) Since 1/cos(x) is the same as sec(x)- the right-hand side- the proof is complete.
What is sin cos tan csc sec cot of 135 degrees?
All those can be calculated quickly with your calculator. Just be sure it is in "degrees" mode (not in radians). Also, use the following identities: csc(x) = 1 / sin(x) sec(x) = 1 / cos(x) cot(x) = 1 / tan(x) or the equivalent cos(x) / sin(x)
How do you apply double angle formulas for csc sec cot?
write in terms of sin, cos or tan then use the double angle formulae. I.e. cosec(x)=1/sin(x) =1/[2sin(x)cos(x)]
How many inches equal a 30 percent incline?
It depends on how high and long the incline is. you need to use cos, sin, and tan to figure it out
How do you solve secant-tangent and tangent-tangent angles?
you solve secant angles when you have the hypotenuse and adjacent sides. sec=1/cos or, cos^-1 (reciprocal identity property) Tangent is solved when you have adjacent and opposite sides, or you can look at it as its what you use when you dont have the hypotenuse. tan=sin/cos or tan=opp/adj or tan=y/x
How do you simplify sec x tan x parenthesees one minus sin squared x?
You use the identity sin2x + cos2x = 1 (to simplify the expression in parentheses), and convert all functions to sines and cosines. sec x tan x (1 - sin2x) = (1/cos x) (sin x / cos x) (cos2x) = (sin x / cos2x) cos2x = sin x
How do you solve double angle equations for trigonometry?
There are two ways to solve for the double angle formulas in trigonometry. The first is to use the angle addition formulas for sine and cosine. * sin(a + b) = sin(a)cos(b) + cos(a)sin(b) * cos(a + b) = cos(a)cos(b) - sin(a)sin(b) if a = b, then * sin(2a) = sin(a)cos(a) + cos(a)sin(a) = 2sin(a)cos(a) * cos(2a) = cos2(a) - sin2(b) The cooler way to solve for the double angle formulas is to use Euler's identity. eix = cos(x) + i*sin(x). Yes, that is "i" as in imaginary number. we we put 2x in for x, we get * e2ix = cos(2x) + i*sin(2x) This is the same as * (eix)2 = cos(2x) + i*sin(2x) We can substitute our original equation back in for eix. * (cos(x) + i*sin(x))2 = cos(2x) + i*sin(2x) We can distribute the squared term. * cos2(x) + i*sin(x)cos(x) + i*sin(x)cos(x) + (i*sin(x))2 = cos(2x) + i*sin(2x) And simplify. Because i is SQRT(-1), the i squared term becomes negative. * cos2(x) + 2i*sin(x)cos(x) - sin2(x) = cos(2x) + i*sin(2x) * cos2(x) - sin2(x) + 2i*sin(x)cos(x) = cos(2x) + i*sin(2x) Now you can plainly see both formulas in the equation arranged quite nicely. I don't yet know how to get rid of the i, but I'm working on it.
How do you use the Euler's formula to obtain the sin3X in terms of cosX and sinX?
Using Euler's Formula, you use (cos(x) + i sin(x))^n = cos (nx) + i sin(nx) Now you let n=3 (cos(x) + i sin (x))3 = cos(3x) + i sin (3x) (cos(x))3 + 3(cos(x))2 * i sin(x) + 3cos(x) * i2 (sin(x))3 = cos(3x)+ i sin(3x) (cos(x))3 + i(3sin(x)(cos (x))2) - 3cos(x)(sin(x)2) - i(sin(x))3 = cos (3x) + i sin(3x) Now only use the terms with i in them to figure out what sin(3x) is... 3sin(x)(cos(x))2 - (sin(x))3 = sin(3x) Hope this helps! :D
