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Yes, it is.
the basic identity is for a double angle relation:
cos 2x = 2 cosx cos x -1
since sec x =1/cos x if we multiply both sides by sec x we get
cos2xsec x = 2cosxcos x/cos x -1/cos x = 2cos x - sec x
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∙ 14y ago
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What is sec theta - 1 over sec theta?
Let 'theta' = A [as 'A' is easier to type] sec A - 1/(sec A) = 1/(cos A) - cos A = (1 - cos^2 A)/(cos A) = (sin^2 A)/(cos A) = (tan A)*(sin A) Then you can swap back the 'A' with theta
What is the integral of sec squared x?
tan(x) + C d/dx tan(x) = d/dx (sin(x))/(cos(x)) = (sin^2(x)+cos^2(x))/(cos^2(x)) = 1/(cos^2(x)) = sec^2(x) NEVER FORGET THE CONSTANT!
Which functions have period 2 pi?
Sin cos sec cosec
Cos 60 equals and sin 60 equals?
cos60= 1/2 sin60=1.732/2
X is acute and sin x equals cos x Find cos?
x = 45 degrees sin(x) = cos(x) = 1/2 sqrt(2)
How do you solve the following identity sec x - cos x equals sin x tan x?
sec x - cos x = (sin x)(tan x) 1/cos x - cos x = Cofunction Identity, sec x = 1/cos x. (1-cos^2 x)/cos x = Subtract the fractions. (sin^2 x)/cos x = Pythagorean Identity, 1-cos^2 x = sin^2 x. sin x (sin x)/(cos x) = Factor out sin x. (sin x)(tan x) = (sin x)(tan x) Cofunction Identity, (sin x)/(cos x) = tan x.
What is the solution to cos equals sec-sintan?
I'm not really sure what you mean by "the solution", but that equation cos = sec - sintan does simplify down to sin^2 + cos^2 = 1 which so happens to be an identity. I'm not sure if that's what you're looking for, but if it is, here are the steps in simplifying it. 1. Convert sec to 1/cos 2. Convert tan into sin/cos and multiply it by sin sintan = sin(sin/cos) = (sin^2)/cos You then have cos = 1/cos - (sin^2/cos) 3. Multiply everything by cos cos^2 = 1 - sin^2 4. And finally, send the sin^2 over to the left side by adding it (since it is being subracted on the right) You should see this sin^2 + cos^2 = 1 which is an identity.
What is 2 by cos a?
2/cos(a) = 2 sec(a)
Find the principal solution of sec x equals 2?
sec(x) = 2 so cos(x) = 1/2 and so x = pi/3
What is sec theta - 1 over sec theta?
Let 'theta' = A [as 'A' is easier to type] sec A - 1/(sec A) = 1/(cos A) - cos A = (1 - cos^2 A)/(cos A) = (sin^2 A)/(cos A) = (tan A)*(sin A) Then you can swap back the 'A' with theta
Is 1- cos 2 x 1 plus cos 2 x equals sin squared x cos squared x an identity?
No, (sinx)^2 + (cosx)^2=1 is though
How do you prove tan x plus tan x sec 2x equals tan 2x?
tan x + (tan x)(sec 2x) = tan 2x work dependently on the left sidetan x + (tan x)(sec 2x); factor out tan x= tan x(1 + sec 2x); sec 2x = 1/cos 2x= tan x(1 + 1/cos 2x); LCD = cos 2x= tan x[cos 2x + 1)/cos 2x]; tan x = sin x/cos x and cos 2x = 1 - 2 sin2 x= (sin x/cos x)[(1 - 2sin2 x + 1)/cos 2x]= (sin x/cos x)[2(1 - sin2 x)/cos 2x]; 1 - sin2 x = cos2 x= (sin x/cos x)[2cos2 x)/cos 2x]; simplify cos x= (2sin x cos x)/cos 2x; 2 sinx cos x = sin 2x= sin 2x/cos 2x= tan 2x
What is the integral of sec squared x?
tan(x) + C d/dx tan(x) = d/dx (sin(x))/(cos(x)) = (sin^2(x)+cos^2(x))/(cos^2(x)) = 1/(cos^2(x)) = sec^2(x) NEVER FORGET THE CONSTANT!
Solve sec2x equals secx plus 2?
You must mean, either,(1) sec2 x = sec x + 2, or(2) sec(2x) = sec x + 2.Let's first assume that you mean:sec2 x = sec x + 2;whence,if we let s = sec x, we have,s2 = s + 2,s2 - s - 2 = (s - 2)(s + 1) = 0, ands = 2 or -1; that is,sec x = 2 or -1.As, by definition,cos x = 1/sec x ,this means thatcos x = ½ or -1.Therefore, providing that the first assumption is correct,x = 60°, 180°, or 300°; or,if you prefer,x = ⅓ π, π, or 1⅔ π.Now, let's assume, instead, that you mean:sec(2x) = sec x + 2;whence,1/(cos(2x) = (1/cos x) + 2.If we let c = cos x,then we have the standard identity,2c2 - 1 = cos (2x); and,thus, it follows that1/(cos(2x) = 1/(2c2 - 1)= (1/c) + 2 = (1 + 2c)/c.This gives,1/(2c2 - 1) = (1 + 2c)/c;(2c2 - 1)(2c + 1) = 4c3 + 2c2 - 2c - 1 = c; and4c3 + 2c2 - 3c - 1 = (c + 1)(4c2 - 2c - 1) = 0.As our concern is only with real roots,c = cos x = -1; and,therefore, providing that the second assumption is correct,x = 180°; or,if you like,x = π.
What is Cos 2 x divided by cos x?
cos 2x = cos2 x - sin2 x = 2 cos2 x - 1; whence, cos 2x / cos x = 2 cos x - (1 / cos x) = 2 cos x - sec x.
Csc squared divided by cot equals csc x sec. can someone make them equal?
cot(x)=1/tan(x)=1/(sin(x)/cos(x))=cos(x)/sin(x) csc(x)=1/sin(x) sec(x)=1/cos(x) Therefore, (csc(x))2/cot(x)=(1/(sin(x))2)/cot(x)=(1/(sin(x))2)/(cos(x)/sin(x))=(1/(sin(x))2)(sin(x)/cos(x))=(1/sin(x))*(1/cos(x))=csc(x)*sec(x)
What is the solution to sec plus tan equals cos over 1 plus sin?
sec + tan = cos /(1 + sin) sec and tan are defined so cos is non-zero. 1/cos + sin/cos = cos/(1 + sin) (1 + sin)/cos = cos/(1 + sin) cross-multiplying, (1 + sin)2 = cos2 (1 + sin)2 = 1 - sin2 1 + 2sin + sin2 = 1 - sin2 2sin2 + 2sin = 0 sin2 + sin = 0 sin(sin + 1) = 0 so sin = 0 or sin = -1 But sin = -1 implies that cos = 0 and cos is non-zero. Therefore sin = 0 or the solutions are k*pi radians where k is an integer.
