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Trig homework help!?

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asked Jul 16, 2014 in TRIGONOMETRY by anonymous

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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.

answered Jul 16, 2014 by lilly Expert
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  • 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)

answered Jul 16, 2014 by lilly Expert

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