Trig homework help!?

Question

Answer 1

10).

• a). csc(tan^{- 1}(√(9 - x²)/x)).

Use the formula : √(1 - x²)/x = tan(cos^{- 1} x).

√(9 - x²)/x = [√(1 - (x/3)²)/(x/3)] = tan(cos^{- 1} (x/3)).

⇒ csc(tan^{- 1}(√(9 - x²)/x)) = csc[ tan^{- 1}(tan(cos^{- 1} (x/3))) ]

= csc(cos^{- 1} (x/3))

= 1/sin(cos^{- 1} (x/3))

Use formula : sin(cos^{- 1} (x)) = √(1 - x²).

= 1/√(1 - (x/3)²)

= 3/√(9 - x²).

Therefore,  csc(tan^{- 1}(√(9 - x²)/x)) = 3/√(9 - x²).

• b). sec(cos^{- 1} [x/√(x² + 5)].

Using reciprocal identity : sec x = 1/cos x.

sec(cos^{- 1} [x/√(x² + 5)] = 1/[ cos(cos^{- 1} [x/√(x² + 5)] ].

cos(cos^-1(x)) = x     for every x in the interval [-1, 1]

cos^-1(cos(x)) = x     for every x in the interval [0, π].

= 1/[x/√(x² + 5)]

= √(x² + 5)/x.

Therefore, sec(cos^{- 1} [x/√(x² + 5)] = √(x² + 5)/x.

Answer 2

• 11). The trigonometric equation is 2cos² x - 7cos x = 3.

Let, cos x = y.

Then, the trigonometric equation is 2y² - 7y - 3 = 0.

2y² - 7y - 3 = 0, is quadratic, use quadratic formula to find the roots of the related quadratic equation.

Solution x = [- b ± √ (b² - 4ac)]/2a.

Compare the equation with standard form of quadratic equation : ax² + bx + c = 0.

a = 2, b = - 7, and c = - 3.

Solution y = [- (- 7) ± √ ((- 7)² - 4*2*(- 3))]/2*2

y = [7 ± √(49 + 24)]/4

y = [7 ± √73]/4.

y = [7 + √73]/4  and  y = [7 - √73]/4

Put, y = cos x.

cos x = [7 + √73]/4  and  cos x = [7 - √73]/4

cos x = cos(cos^{- 1}([7 + √73]/4)) and  cos x = cos(cos^{- 1}([7 - √73]/4))

⇒ x = cos^{- 1}([7 + √73]/4) and  x = cos^{- 1}([7 - √73]/4).

The solution of the given equation is x = cos^{- 1}([7 + √73]/4) and  x = cos^{- 1}([7 - √73]/4) in the interval [0, 2π).

• 12). The trigonometric equation is cos (x/2) - cos x = 1.

Half - angle formula : cos(x/2) = ±√[(1 + cos x)/2].

±√[(1 + cos x)/2] - cos x = 1

±√[(1 + cos x)/2] = 1 + cos x

Squaring on each side.

( ±√[(1 + cos x)/2] )² = (1 + cos x)²

(1 + cos x)/2 = 1 + 2cos x + cos² x

1 + cos x = 2 + 4cos x + 2cos² x

2 + 4cos x - cos x + 2cos² x - 1 = 0

2cos² x  + 3cos x + 1= 0

2cos² x  + 2cos x + cos x + 1= 0

2cos x(cos x + 1) + 1(cos x + 1)= 0

(cos x + 1)(2cos x + 1) = 0

⇒ cos x + 1 = 0  and  2cos x + 1 = 0

cos x = - 1  and  cos x = - 1/2

cos x = cos π  and  cos x = cos(2π/3) in the interval [0, 2π).

⇒ x = π and x = 2π/3.

The solutions of the given equations are x = π and x = 2π/3 in the interval [0, 2π)