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Q: Cos x sin x identity

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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.

Sine sum identity: sin (x + y) = (sin x)(cos y) + (cos x)(sin y)Sine difference identity: sin (x - y) = (sin x)(cos y) - (cos x)(sin y)Cosine sum identity: cos (x + y) = (cos x)(cos y) - (sin x)(sin y)Cosine difference identity: cos (x - y) = (cos x)(cos y) + (sin x)(sin y)Tangent sum identity: tan (x + y) = [(tan x) + (tan y)]/[1 - (tan x)(tan y)]Tangent difference identity: tan (x - y) = [(tan x) - (tan y)]/[1 + (tan x)(tan y)]

Yes. sin(A+B) = sin A cos B + cos A sin B If A = B = x, this becomes: sin(x+x) = sin x cos x + cos x sin x → sin 2x = 2 sin x cos x

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.

(sin(x)cot(x) - cos(x))/tan(x)(Multiply by tan(x)/tan(x))sin(x) - cos(x)tan(x)(tan(x) = sin(x)/cos(x))sinx - cos(x)(sin(x)/cos(x))(cos(x) cancels out)sin(x) - sin(x)0

An identity is a statement which says two quantities are equal, like as x + y = y + x or sin (x + y ) = sin x cos y + cos x sin y .

(2 sin^2 x - 1)/(sin x - cos x) = sin x + cos x (sin^2 x + sin^2 x - 1)/(sin x - cos x) =? sin x + cos x [sin^2 x - (1 - sin^2 x)]/(sin x - cos x) =? sin x + cos x (sin^2 x - cos^2 x)/(sin x - cos x) =? sin x + cos x [(sin x - cos x)(sin x + cos x)]/(sin x - cos x) =? sin x + cos x sin x + cos x = sin x + cos x

One relationship is: cos(x) = sin(90Â° - x) if you use degrees. Or in radians: cos(x) = sin(pi/2 - x) Another relationship is the pythagorean identity.

(tan x - sin x)/(tan x sin x) = (tan x sin x)/(tan x + sin x)[sin x/cos x) - sin x]/[(sin x/cos x)sin x] =? [(sin x/cos x)sin x]/[sin x/cos x) + sin x][(sin x - sin x cos x)/cos x]/(sin2 x/cos x) =? (sin2 x/cos x)/[(sin x + sin x cos x)/cos x)(sin x - sin x cos x)/sin2 x =? sin2 x/(sin x + sin x cos x)[sin x(1 - cos x)]/sin2 x =? sin2 x/[sin x(1 + cos x)(1 - cos x)/sin x =? sin x/(1 + cos x)(1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[(1 + cos x)(1 - cos x)](1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[1 - cos2 x)(1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[1 - (1 - sin2 x)](1 - cos x)/sin x =? [(sin x)(1 - cos x)]/sin2 x(1 - cos x)/sin x = (1 - cos x)/sin x True

The easiest way to approach this problem is by rewriting the left hand side entirely in terms of sin and cos and then simplifying. To do so, use the fact that cot(x)=cos(x)/sin(x) to get that 2*cot(x)*sin(x)*cos(x)=2*cos(x)/sin(x)*sin(x)*cos(x)=2*cos(x)² Next, we will try to simplify the right hand side by factoring and utilizing the formula cos(x)²+sin(x)²=1 which implies that 1-sin(x)²=cos(x)² 2-2sin(x)²=2*(1-sin(x)²)=2*cos(x)² Since both sides can be simplified to equal the same thing, both sides must always be equal, and the equation 2*cot(x)*sin(x)*cos(x)=2-2sin(x)² must be an identity

cot x = (cos x) / (sin x) cos (x - 180) = cos x cos 180 + sin x sin 180 = - cos x sin (x - 180) = sin x cos 180 - cos x sin 180 = - sin x cot (x - 180) = (cos (x - 180)) / (sin (x - 180)) = (- cos x) / (- sin x) = (cos x) / (sin x) = cot x

∫cos2(x).dxUse the identity cos2(x) = (1/2)(1+cos(2x))∫(1/2)(1+cos(2x))dxPull out constant:(1/2)∫(1+cos(2x))dxIntegrate:(1/2)(x + sin(2x)/2) + CSimplify:x/2 + sin(2x)/4 + CThe identity sin(2x) = 2sin(x)cos(x) can be used to rewrite it as(x + sin(x)cos(x))/2 + C

sin x/(1+cos x) + cos x / sin x Multiply by sin x (1+cos x) =[(sin^2 x + cos x(1+cos x) ] / sin x (1+cos x) = [(sin^2 x + cos x + cos^2 x) ] / sin x (1+cos x) sin^2 x + cos^2 x = 1 = (1+cos x) / sin x (1+cos x) = 1/sin x

cos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan x

First convert everything to sines and cosines:sin x + sin x cos x / sin x = 1 / sin xsin x + cos x = 1 / sin xMultiplying by sin x:sin2x + sin x cos x = 1Using the identity sin2 + cos2x = 1:sin2x + sin x cos x = sin2x + cos2xsin x cos x = cos2xDividing by cos x:sin x = cos xThe solution is therefore x = pi / 4 radians, or x = 5 pi / 4 radians.The division by cos x assumed that cos x was not equal to zero; this possibility must be explored in the original equation. When cos x = 0, sin x = 1 or -1, and the angle x = pi/2 or -pi/2. It seems both of these are solutions, too.First convert everything to sines and cosines:sin x + sin x cos x / sin x = 1 / sin xsin x + cos x = 1 / sin xMultiplying by sin x:sin2x + sin x cos x = 1Using the identity sin2 + cos2x = 1:sin2x + sin x cos x = sin2x + cos2xsin x cos x = cos2xDividing by cos x:sin x = cos xThe solution is therefore x = pi / 4 radians, or x = 5 pi / 4 radians.The division by cos x assumed that cos x was not equal to zero; this possibility must be explored in the original equation. When cos x = 0, sin x = 1 or -1, and the angle x = pi/2 or -pi/2. It seems both of these are solutions, too.First convert everything to sines and cosines:sin x + sin x cos x / sin x = 1 / sin xsin x + cos x = 1 / sin xMultiplying by sin x:sin2x + sin x cos x = 1Using the identity sin2 + cos2x = 1:sin2x + sin x cos x = sin2x + cos2xsin x cos x = cos2xDividing by cos x:sin x = cos xThe solution is therefore x = pi / 4 radians, or x = 5 pi / 4 radians.The division by cos x assumed that cos x was not equal to zero; this possibility must be explored in the original equation. When cos x = 0, sin x = 1 or -1, and the angle x = pi/2 or -pi/2. It seems both of these are solutions, too.First convert everything to sines and cosines:sin x + sin x cos x / sin x = 1 / sin xsin x + cos x = 1 / sin xMultiplying by sin x:sin2x + sin x cos x = 1Using the identity sin2 + cos2x = 1:sin2x + sin x cos x = sin2x + cos2xsin x cos x = cos2xDividing by cos x:sin x = cos xThe solution is therefore x = pi / 4 radians, or x = 5 pi / 4 radians.The division by cos x assumed that cos x was not equal to zero; this possibility must be explored in the original equation. When cos x = 0, sin x = 1 or -1, and the angle x = pi/2 or -pi/2. It seems both of these are solutions, too.

The question is ambiguous: does it refer to 1/sin(x) + cos(x) or to 1/[sin(x)+cos(x)]?The question is ambiguous: does it refer to 1/sin(x) + cos(x) or to 1/[sin(x)+cos(x)]?The question is ambiguous: does it refer to 1/sin(x) + cos(x) or to 1/[sin(x)+cos(x)]?The question is ambiguous: does it refer to 1/sin(x) + cos(x) or to 1/[sin(x)+cos(x)]?

You need to know the trigonometric formulae for sin and cos of compound angles. sin(x+y) = sin(x)*cos(y)+cos(x)*sin(y) and cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y) Using these, y = x implies that sin(2x) = sin(x+x) = 2*sin(x)cos(x) and cos(2x) = cos(x+x) = cos^2(x) - sin^2(x) Next, the triple angle formulae are: sin(3x) = sin(2x + x) = 3*sin(x) - 4*sin^3(x) and cos(3x) = 4*cos^3(x) - 3*cos(x) Then the left hand side = 2*[3*sin(x) - 4*sin^3(x)]/sin(x) + 2*[4*cos^3(x) - 3*cos(x)]/cos(x) = 6 - 8*sin^2(x) + 8cos^2(x) - 6 = 8*[cos^2(x) - sin^2(x)] = 8*cos(2x) = right hand side.

(cos x sin x) / (cos x sin x) = 1. The derivative of a constant, such as 1, is zero.

You can't. tan x = sin x/cos x So sin x tan x = sin x (sin x/cos x) = sin^2 x/cos x.

It isn't. The derivate of sin x = cos x.It isn't. The derivate of sin x = cos x.It isn't. The derivate of sin x = cos x.It isn't. The derivate of sin x = cos x.

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.

(tan x + cot x)/sec x . csc x The key to solve this question is to turn tan x, cot x, sec x, csc x into the simpler form. Remember that tan x = sin x / cos x, cot x = 1/tan x, sec x = 1/cos x, csc x = 1/sin x The solution is: [(sin x / cos x)+(cos x / sin x)] / (1/cos x . 1/sin x) [(sin x . sin x + cos x . cos x) / (sin x . cos x)] (1/sin x cos x) [(sin x . sin x + cos x . cos x) / (sin x . cos x)] (sin x . cos x) then sin x. sin x + cos x . cos x sin2x+cos2x =1 The answer is 1.

== cot(x)== 1/tan(x) = cos(x)/sin(x) Now substitute cos(x)/sin(x) into the expression, in place of cot(x) So now: sin(x) cot(x) cos(x) = sin(x) cos(x) (cos(x)/sin(x) ) sin(x) cos(x) cos(x)/sin(x) The two sin(x) cancel, leaving you with cos(x) cos(x) Which is the same as cos2(x) So: sin(x) cot(x) cos(x) = cos2(x) ===

2

To simplify this sort of things, it helps if, first of all, you convert everything to sines and cosines.cos x cot x + tan x (original equation)= cos (cos x / sin x) + (sin x / cos x) (convert to sin and cos)= cos2x / sin x + sin x / cos x (multiplying in the first term)= (sin x cos2x + sin x cos x) / sin x cos x (converting common denominator)= (sin x cos x) (cos x + 1) / (sin x cos x) (factoring the numerator)= cos x + 1 (cancelling factors in numerator and denominator)