4 trigonometric calculus Q

1) y = sin x + 1 - 2 sin2 x

= -2[sin2 x - (1/2) sin x] + 1

= -2[sin2 x - (1/2) sin x + 1/16] + 1 + 1/8

= -2(sin x - 1/4)2 + 9/8

For y to be min., sin x = 1/4

Hence first occurence is x = 0.253

For y to be max., sin x = -1

Hence first occurence is x = 3π/2

2) dy/dx = cos x sin 2x + 2 sin x cos 2x

= 2 sin x cos2 x + 2 sin x (2 cos2 x - 1)

= 2 sin x (3 cos2 x - 1)

When x = 0, dy/dx = 0

When x < 0 for a small amount. dy/dx < 0

When x > 0 for a small amount. dy/dx > 0

So x = 0 is the first occurence of min.

When cos x = 1/√3, dy/dx = 0, with first occurence at x = 0.955

When x < 0.955 for a small amount. dy/dx > 0

When x > 0.955 for a small amount. dy/dx < 0

So x = 0.955 is the first occurence of max.

3) y = (tan x - 1)2 - 1

So first occurence of min. is at x = π/4 and first occurence of max. is at x = π/2

4) a sin x + b cos x

= √(a2 + b2) [a sin x/√(a2 + b2) + b cos x/√(a2 + b2)]

= √(a2 + b2) sin (x + θ) where θ = tan-1 (b/a)

So the expression has a max. of √(a2 + b2) and min. of -√(a2 + b2)

2011-08-17 17:44:36 補充：
Pls refer to the diagram below:
http://i1191.photobucket.com/albums/z467/robert1973/Aug11/Crazytrigo1.jpg

We are going to find θ such that:

a sin x + b cos x = √(a^2 + b^2) sin (x + θ)

2011-08-17 21:38:57 補充：
This is called the subsudiary angle form, for any a sin x + b cos x, we have to first extract:

√(a^2 + b^2)

So that in the bracket, a/√(a^2 + b^2) and b/√(a^2 + b^2) are there.

2011-08-17 21:39:02 補充：
Since [a/√(a^2 + b^2)]^2 + [b/√(a^2 + b^2)]^2 = 1, we can be sure that we can find a θ such that:

cos θ = a/√(a^2 + b^2)

sin θ = b/√(a^2 + b^2)

2011-08-18 14:45:39 補充：
a sin x + b cos x = √(a^2 + b^2) [a sin x/√(a^2 + b^2) + b cos x/√(a^2 + b^2)]
= √(a^2 + b^2) (sin x cos θ + cos x sin θ)
= √(a^2 + b^2) sin (x + θ)

2011-08-18 20:42:02 補充：
Yes, finally you get it!!!