What is t for cost+1=2sint+sin2t ?

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To determine the angle t, we'll have to solve the equation. We notice that one of the terms of the identity is the function sine of a double angle.

We'll apply the formula for the double angle:

sin 2a = sin (a+a)=sina*cosa + sina*cosa=2sina*cosa

We'll replace 2a by 2t and we'll get:

sin 2t = 2sin t*cos t

We'll re-write the equation, moving all terms to one side:

2sin t*cos t + 2sint - cost - 1 = 0

We'll factorize by 2sin t the first 2 terms:

2sint(cos t + 1) - (cos t + 1) = 0

We'll factorize by (cos t + 1):

(cos t + 1)(2sin t - 1) = 0

We'll set each factor as zero:

cos t + 1 = 0

We'll add -1 both sides:

cos t = -1

t = arccos (-1)

t = pi

2sin t - 1 = 0

We'll add 1 both sides:

2sin t = 1

sin t = 1/2

t = arcsin (1/2)

t = pi/6

t = 5pi/6

The angle t has the following values: {pi/6 ; 5pi/6 ; pi}.

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

cost+1 = 2sint +sin2t. We  proceed to solve for t.

We know  sin2t = 2sint cost.

Therefore , the given equation becomes:

cost+1 = 2sint +2sint*cost.

cost+1 = 2sint(1+cost).

Subtract cost+1 from both sides:

0 = 2sint(1+cost) - (1+cost).

0 = 2sint(1+cost) - 1(1+cost).

0 = (cost+1)(2sint-1).

We equate the factors to zero.

cost+1 = 0. Or 2sint -1 = 0.

cost = -1, t = pi or t = (2n+1)pi

2sint = 1,  sint = 1/2. t = npi+(-1)^n*pi/6.

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