Why is the trigonometric equation Sin(2Ө) = 2Sin(Ө) NOT an identity?

Question

I need to explain the identities' truth by referring to the unit circle and the definitions of the various trigonometric functions.

Answer 1

ok ...
draw a triangle with angles 30, 60, 90
sin 30 = 1/2 so our sides are 1, sqrt(3), 2
sin 60 = sqrt(3)/2 and this is not equal to 2/2 =1
therefore its not an identity

in fact sin2@ = 2sin@cos@ = 2sin@ is only satisfied when
sin@ = 0 or cos@=1, i.e. when @ = 2pi*integer

Answer 2

Well since the equation doesn't hold for all values of Ө, it is not an identity. For example, let Ө = 30°. Then
LHS=sin(60°)=sqrt(3)/2
RHS=2*sin(30°)=2*(1/2)=1

Because we've found a counter example, the above equation is not an identity.

Answer 3

Because it doesn't hold for all theta. There a number of possible counter examples. If you recall, the correct identity is

Sin(2Ө) = 2Sin(Ө)Cos(Ө)

So Sin(2Ө) = 2 Sin(Ө) only if Cos(Ө) = 1, which is true only for
Ө = 2 k pi, for some integer k.

Answer 4

An identity must be always true. Use your unit circle to consider a case where your equation fails. Consider the case of 90 degrees. Since the line at 90 degrees is straight up, your sinӨ is equal to 1. If you double the angle, the vertical height (the sine) goes back to zero so doubling of sinӨ (1.0 x 2) is obviously not zero.

Answer 5

If it was an identity it would be true for all values of theta.

Consider theta = pi/2 radians. Then sin(pi)=0 and 2sin(pi/2) = 2. so clearly it doesn't apply to that value .
Probably only true for theta = o , pi, .....npi where n is an integer value.

Answer 6

You only need one false value to show it's not an identity.
Take θ = π/2. Then sin 2θ = sin π = , but
2 sin θ = 2 sin π/2 = 2.
Since we have found a value of θ for which the
equation doesn't hold, it's not an identity.

Answer 7

Hehehe. Since it is *not* an identity, you'll play merry hell trying to explain its 'truth' ☺

Probably the easiest way is to go through the derivation of the formula
sin(A+B) = sin(A)cos(B) + sin(B)cos(A) (which will be in any elementary trig book) and then substitute to get
sin(Φ+Φ) = 2sin(Φ)cos(Φ) which is not equal to 2sin(Φ)

HTH

Doug

Answer 8

Well, firstly, because it's wrong. sine 2 theta does not equal 2 sine theta.

It's not even a valid mathematical expression. So it's not anything at all.

Answer 9

i) By property sec(A+B) = 1/{cos(A+B)}, provided (A+B) ≠ π/2 [Since, cos(π/2) = 0 and sec*cos = 1] ii) ==> sec(A+B) = 1/[cos(A)*cos(B) - sin(A)*sin(B)] {Identity cos(A+B) = cos(A)*cos(B) - sin(A)*sin(B)} iii) Dividing both numerator and denominator by cos(A)*cos(B), the above simplifies to: sec(A+B) = {sec(A)*sec(B)}/{1 - tan(A)*tan(B)} [Proved] Thus sec(A+B) = {sec(A)*sec(B)}/{1 - tan(A)*tan(B)} is an identity