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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 simplify cos theta times csc theta divided by tan theta?
'csc' = 1/sin'tan' = sin/cosSo it must follow that(cos) (csc) / (tan) = (cos) (1/sin)/(sin/cos) = (cos) (1/sin) (cos/sin) = (cos/sin)2
Verify the identity sinx cotx - cosx divided by tanx equals 0?
(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
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 the identity for tan theta?
The identity for tan(theta) is sin(theta)/cos(theta).
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.
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.
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
What is the half angle formula to find the exact value for tan 165?
tan u/2 = sin u/1+cos u
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
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).
Sin x Tan x equals Sin x?
No. Tan(x)=Sin(x)/Cos(x) Sin(x)Tan(x)=Sin2(x)/Cos(x) Cos(x)Tan(x)=Sin(x)
How do you prove sin x tan x equals cos x?
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.
How do you simplify cos theta times csc theta divided by tan theta?
'csc' = 1/sin'tan' = sin/cosSo it must follow that(cos) (csc) / (tan) = (cos) (1/sin)/(sin/cos) = (cos) (1/sin) (cos/sin) = (cos/sin)2
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 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.
