Question

1. Find the exact value of sin( a−B), given that sin a=−4/5 and cos B=12/13, with a in quadrant III and B in quadrant IV.

2. Find all real numbers in the interval [0,2pi) that satisfy the equation.

3sec^2 x tan x =4tan x

3. Simplify the following trigonometric expressions, using identities as needed:

sin(x)/1−cos(x) + 1−cos(x)/sin(x)

Answer 1

(1) Recall that

sin(x - y) = sin(x) cos(y) - cos(x) sin(y)

sin²(x) + cos²(x) = 1

Given that α lies in the third quadrant, and β lies in the fourth quadrant, we expect to have

• sin(α) < 0 and cos(α) < 0

• sin(β) < 0 and cos(β) > 0

Solve for cos(α) and sin(β) :

cos(α) = -√(1 - sin²(α)) = -3/5

sin(β) = -√(1 - cos²(β)) = -5/13

Then

sin(α - β) = sin(α) cos(β) - cos(α) sin(β) = (-4/5) (12/13) - (-3/5) (-5/13)

==> sin(α - β) = -63/65

(2) In the second identity listed above, multiplying through both sides by 1/cos²(x) gives another identity,

sin²(x)/cos²(x) + cos²(x)/cos²(x) = 1/cos²(x)

==> tan²(x) + 1 = sec²(x)

Rewrite the equation as

3 sec²(x) tan(x) = 4 tan(x)

3 (tan²(x) + 1) tan(x) = 4 tan(x)

3 tan³(x) + 3 tan(x) = 4 tan(x)

3 tan³(x) - tan(x) = 0

tan(x) (3 tan²(x) - 1) = 0

Solve for x :

tan(x) = 0 or 3 tan²(x) - 1 = 0

tan(x) = 0 or tan²(x) = 1/3

tan(x) = 0 or tan(x) = ±√(1/3)

x = arctan(0) + nπ or x = arctan(1/√3) + nπ or x = arctan(-1/√3) + nπ

x = nπ or x = π/6 + nπ or x = -π/6 + nπ

where n is any integer. In the interval [0, 2π), we get the solutions

x = 0, π/6, 5π/6, π, 7π/6, 11π/6

(3) You only need to rewrite the first term:

`(\sin(x))/(1-\cos(x)) * (1+\cos(x))/(1+\cos(x)) = (\sin(x)(1+\cos(x)))/(1-\cos^2(x)) = (\sin(x)(1+\cos(x))/(\sin^2(x)) = (1+\cos(x))/(\sin(x))`

Then

`(\sin(x))/(1-\cos(x))+(1-\cos(x))/(\sin(x)) = (1+\cos(x)+1-\cos(x))/(\sin(x))=\frac2{\sin(x)}`