Thread: Graph question :S

The question is:
9(i) Sketch, on a single diagram showing values of x from -180degs to +180degs, the graph of y=tanx and y= 4cos x.

The equation tan x = 4cos x

has two roots in the interval -180degs\< +\<180degs. These are denoted by a(alpha) and b(pheta), where a<b

(ii) Show a and b on your sketch, and express b in terms of a.

(iii) Show that the equation tan x = 4cos x may be written as 4sin^2 x + sin x - 4 =0

Hence find the value of b-a, correct to the nearest degree.

If you can help me in any way possible, i have drawn the graph!

Originally Posted by LoveDeathCab

(iii) Show that the equation tan x = 4cos x may be written as 4sin^2 x + sin x - 4 =0

This has already been answered in a previous post.

I know, its the rest that i am struggling with.

Hello, LoveDeathCab!

9(i) Sketch, on one diagram, the graphs: .y = tan(x) and y = 4·cos(x), -180° < x < +180°

The equation: tan(x) = 4·cos(x) has two roots in the interval [-180°, +180°]
These are denoted by α and β, where α < β.

We see that the graphs intersect on [0°, 90°] and [90°, 180°]

(ii) Show α and β on your sketch, and express β in terms of α.

Are they kidding? . . . We need part (iii) first, don't we? .**

(iii) Show that the equation: tan(x) = 4·cos(x) may be written as: 4·sin²(x) + sin(x) - 4 .= .0

Hence find the value of β - α, correct to the nearest degree.

So that's where that equation came from!


We have a quadratic: .4·sin²(x) + sin(x) - 4 .= .0
. . . . . . . . . . . . . . . . . . . . . . . . . . __
. . . . . . . . . . . . . . . . . . . . . .-1 ± √65
Quadratic Formula: . sin(x) .= .-----------
. . . . . . . . . . . . . . . . . . . . . . . . 8

. . . . . . . . . . . . . . . . . . . __
We have: .sin(x) .= .(-1 - √65)/8 .= .-1.132782219 ... no real roots

. . . . . . . . . . . . . . . . . . . . . . .__
And we have: .sin(x) .= .(-1 + √65)/8 .= .0.882782219
. . Hence: .x .≈ .62°, 118°


Therefore: .β - α .= .118° - 62° .= .56°

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

Even if we did part (iii) first, it's still pretty awful . . .

The equation: .sin²(x) + ¼·sin(x) - 1 .= .0 has two roots: α and β.

. . Then: .sin(α) + sin(β) .= .-¼

Solve for β: .β .= .arcsin(-sinα - ¼)