1. A cosine can be calculated as the length of the adjacent side divided by the length of the hypotenuse of a right angled triangle. The tangent measure is the length of the opposite side divided by the length of the adjacent side. As the cosx = 0.6 = 3/5 then we are dealing with a right angled triangle where the adjacent side measures 3 units and the hypotenuse 5 units - thus the opposite side measures 4 units as by Pythagoras' Theorem, 5² = 3² + 4². Then tanx = 4/3.
2. sin²x+cos²x=1.
sin² x=1-0.6²=0.64 and sin x=0.8.
tan x=sin x/cos x=0.8/0.6=8/6=4/3.
To solve this question, we must first solve for x. We can do that by taking the inverse cosine of 0.6, which is 53.13010235. Using this x, we can solve for tan x by inputting tan 53.13010235 in a calculator, which gives 1.33333333.
sin, tan and cos can be defined as functions of an angle. But they are not functions of a triangle - whether it is a right angled triangle or not.
A useful property in Trigonometry is: tan(x) = sin(x) / cos(x) So, cos(x) tan(x) = cos(x) [ sin(x) / cos (x)] = sin(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)]
sine(sin) = opp/hypcosecant(q) = hyp/oppcosine(cos) = adj/hypsecant(q) = hyp/adjtangent(tan) = opp/adjcotangent(q) = adj/opp
tan x = sin x / cos x, so:lim (tan x / x) = lim (sin x / x cos x). Since it is known that the limit of sin x / x = 1, you have lim 1 / cos x = 1 (since cos 0 = 1).tan x = sin x / cos x, so:lim (tan x / x) = lim (sin x / x cos x). Since it is known that the limit of sin x / x = 1, you have lim 1 / cos x = 1 (since cos 0 = 1).tan x = sin x / cos x, so:lim (tan x / x) = lim (sin x / x cos x). Since it is known that the limit of sin x / x = 1, you have lim 1 / cos x = 1 (since cos 0 = 1).tan x = sin x / cos x, so:lim (tan x / x) = lim (sin x / x cos x). Since it is known that the limit of sin x / x = 1, you have lim 1 / cos x = 1 (since cos 0 = 1).
There are many. For example, if A and B are the two acute angles, then A + B = 90 degrees or sin(A) = cos(B) or cos(A) = sin(B) or tan(A) = 1/tan(B)
3cos(y) = 3/(sqrt(1+x^2)
To show that (cos tan = sin) ??? Remember that tan = (sin/cos) When you substitute it for tan, cos tan = cos (sin/cos) = sin QED
If ø is an obtuse angle then (180 - ø) is an acute angle and: sin ø = sin (180 - ø) cos ø = -cos (180 - ø) tan ø = -tan (180 - ø)
sin, tan and cos can be defined as functions of an angle. But they are not functions of a triangle - whether it is a right angled triangle or not.
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
The definition of tan(x) = sin(x)/cos(x). By this property, cos(x)tan(x) = sin(x).
A useful property in Trigonometry is: tan(x) = sin(x) / cos(x) So, cos(x) tan(x) = cos(x) [ sin(x) / cos (x)] = sin(x)
Remember that tan = sin/cos. So your expression is sin/cos times cos. That's sin(theta).
When tan A = 815, sin A = 0.9999992 and cos A = 0.0012270 so that sin A + cos A*cos A*(1-cos A) = 1.00000075, approx.
If tan 3a is equal to sin cos 45 plus sin 30, then the value of a = 0.4.
tan(9) + tan(81) = sin(9)/cos(9) + sin(81)/cos(81)= {sin(9)*cos(81) + sin(81)*cos(9)} / {cos(9)*cos(81)} = 1/2*{sin(-72) + sin(90)} + 1/2*{sin(72) + sin(90)} / 1/2*{cos(-72) + cos(90)} = 1/2*{sin(-72) + 1 + sin(72) + 1} / 1/2*{cos(-72) + 0} = 2/cos(72) since sin(-72) = -sin(72), and cos(-72) = cos(72) . . . . . (A) Also tan(27) + tan(63) = sin(27)/cos(27) + sin(63)/cos(63) = {sin(27)*cos(63) + sin(63)*cos(27)} / {cos(27)*cos(63)} = 1/2*{sin(-36) + sin(90)} + 1/2*{sin(72) + sin(36)} / 1/2*{cos(-36) + cos(90)} = 1/2*{sin(-36) + 1 + sin(36) + 1} / 1/2*{cos(-36) + 0} = 2/cos(36) since sin(-36) = -sin(36), and cos(-36) = cos(36) . . . . . (B) Therefore, by (A) and (B), tan(9) - tan(27) - tan(63) + tan(81) = tan(9) + tan(81) - tan(27) - tan(63) = 2/cos(72) – 2/cos(36) = 2*{cos(36) – cos(72)} / {cos(72)*cos(36)} = 2*2*sin(54)*sin(18)/{cos(72)*cos(36)} . . . . . . . (C) But cos(72) = sin(90-72) = sin(18) so that sin(18)/cos(72) = 1 and cos(36) = sin(90-36) = sin(54) so that sin(54)/cos(36) = 1 and therefore from C, tan(9) – tan(27) – tan(63) + tan(81) = 2*2*1*1 = 4