cos(x)=sin(x-tau/4) tan(x)=sin(x)/cos(x) sin(x)=tan(x)*cos(x) cos(x)=tan(x-tau/4)*cos(x-tau/4) you can see that we have some circular reasoning going on, so the best we can do is express it as a combination of sines and cotangents: cos(x)=1/cot(x-tau/4)*sin(x-tau/2) tau=2*pi
csc(x) = 1/sin(x) = +/- 1/sqrt(1-cos^2(x))
The best way to answer this question is with the angle addition formulas. Sin(a + b) = sin(a)cos(b) + cos(a)sin(b) and cos(a + b) = cos(a)cos(b) - sin(a)sin(b). If you compute this repeatedly until you get sin(3x)cos(4x) = 3sin(x) - 28sin^3(x) + 56sin^5(x) - 32sin^7(x).
It helps to convert this kind of equation into one that has only sines and cosines, by using the basic definitions of the other functions in terms of sines and cosines. sin x / (1 - cos x) = csc x + cot x sin x / (1 - cos x) = 1 / sin x + cos x / sin x Now it should be easy to do some simplifications: sin x / (1 - cos x) = (1 + cos x) / sin x Multiply both sides by 1 + cos x: sin x (1 + cos) / ((1 - cos x)(1 + cos x)) = (1 + cos x)2 / sin x sin x (1 + cos) / (1 - cos2x) = (1 + cos x)2 / sin x sin x (1 + cos) / sin2x = (1 + cos x)2 / sin x sin x (1 + cos x) / sin x = (1 + cos x)2 1 + cos x = (1 + cos x)2 1 = 1 + cos x cos x = 0 So, cos x can be pi/2, 3 pi / 2, etc. In some of the simplifications, I divided by a factor that might be equal to zero; this has to be considered separately. For example, what if sin x = 0? Check whether this is a solution to the original equation.
cos(30)cos(55)+sin(30)sin(55)=cos(30-55) = cos(-25)=cos(25) Note: cos(a)=cos(-a) for any angle 'a'. cos(a)cos(b)+sin(a)sin(b)=cos(a-b) for any 'a' and 'b'.
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.
There are six trigonometric ratios. Although applicable for any angle, they are usually introduced in the context of a right angled triangle. The full names of the main three ratios are sine, cosine, tangent. The other three ratios are reciprocals, which are cosecant, secant and cotangent, respectively.Suppose ABC is a triangle which is right angled at B. Thus AC is the hypotenuse.sin(A) = BC/AC = cos(C)cos(A) = AB/AC = sin(C)tan(A) = BC/AB
There are three trigonometrical ratios for finding the angles and lengths of a right angled triangle and they are tangent, cosine and sine usually abbreviated to tan, cos and sin respectively. tan = opp/adj cos = adj/hyp sin = opp/hyp Note that: opp, adj and hyp are abbreviations for opposite, adjacent and hypotenuse sides of a right angled triangle respectively.
sin = sqrt(1 - cos^2)tan = sqrt(1 - cos^2)/cossec = 1/coscosec = 1/sqrt(1 - cos^2)cot = cos/sqrt(1 - cos^2)
sin2(1) = 1 - cos2(1) = 1 - [cos(1)]2
(sin x + cos x) / cosx = sin x / cos x + cosx / cos x = tan x + 1
sin, cos and tan
Cos(-155) = cos(155) = 1 - cos(180-155) = 1-cos(25).
cos(x)=sin(x-tau/4) tan(x)=sin(x)/cos(x) sin(x)=tan(x)*cos(x) cos(x)=tan(x-tau/4)*cos(x-tau/4) you can see that we have some circular reasoning going on, so the best we can do is express it as a combination of sines and cotangents: cos(x)=1/cot(x-tau/4)*sin(x-tau/2) tau=2*pi
csc(x) = 1/sin(x) = +/- 1/sqrt(1-cos^2(x))
Secant is a trignometric function. In a right triangle, the secant of an angle is the hypotenuse over the adjacent side. It is also the inverse of cosine. For example secant(x) = 1/cos(x)
To simplify such expressions, it helps to express all trigonometric functions in terms of sines and cosines. That is, convert tan, cot, sec or csc to their equivalent in terms of sin and cos.