0
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.
Add your answer:
Earn +20 pts
How would you find out the sin tan and cos of right angle triangle?
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.
What is cos x tan x simplified?
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)
What are the sum and difference identities for the sine cosine and tangent functions?
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)]
How do you find sin cos and tan?
sine(sin) = opp/hypcosecant(q) = hyp/oppcosine(cos) = adj/hypsecant(q) = hyp/adjtangent(tan) = opp/adjcotangent(q) = adj/opp
Limit when x goes to 0 tanx divided by x?
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).
What is the relationship between the acute angles of a right triangle?
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)
If tan y equals to x where y is acute find 3 cos y in terms of x?
3cos(y) = 3/(sqrt(1+x^2)
How do you verify the identity of cos θ tan θ equals sin θ?
To show that (cos tan = sin) ??? Remember that tan = (sin/cos) When you substitute it for tan, cos tan = cos (sin/cos) = sin QED
How do you calculate trigonometric ratios of obtuse angles?
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)
How would you find out the sin tan and cos of right angle triangle?
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.
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 cos x tan x simlpified?
The definition of tan(x) = sin(x)/cos(x). By this property, cos(x)tan(x) = sin(x).
What is cos x tan x simplified?
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)
How do you simplify tan theta cos theta?
Remember that tan = sin/cos. So your expression is sin/cos times cos. That's sin(theta).
What is the value of sin A plus cos Acos A (1-cos A) when tan A 815?
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.
Find the value of a if tan 3a is equal to sin cos 45 plus sin 30?
If tan 3a is equal to sin cos 45 plus sin 30, then the value of a = 0.4.
How tan9-tan27-tan63 tan81 equals 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
