IIT JEE , AIEEE Forum - Mathematics , Trignometry

PRIYAMZ

PROVE THAT :

-2<=cosx(sinx+

(sin²x+3))<=2

priyamz

avinash.sharma

we know

0<= SinX <= 1 when 0<= X <= 90

1<= SinX <= 0 when 90<= X <= 180

0<= SinX <= -1 when 180<= X <= 270

-1<= SinX <= 0 when 270<= X <= 360

and

1<= CosX <= 0 when 0<= X <= 90

0<= CosX <= -1 when 90<= X <= 180

-1<= CosX <= 0 when 180<= X <= 270

0<= CosX <= 1 when 270<= X <= 360

So

-1 <= SinX <= 1 when 0<= X <= 360

0 <= Sin²X <= 1

3 <= Sin²X+3 <= 4

-2<= Ö (Sin²X+3) <= 2 [ as Ö 4 = ± 2 and Ö 3 = ± 1.7320 ]

-3 <= SinX + Ö (Sin²X+3) <= 3

Now we have to pay proper attention for the movement in SinX + Ö (Sin²X+3) and CosX with respect to changes in X so that we can combine these two independent functions. It is very important step of this solution. Lets take f(x) as CosX and g(X)= SinX + Ö (Sin²X+3)

g(X) moves 2 to 3 when X moves from 15 to 90 and same time f(x) moves from 0.966 to 0, so f(x).g(x) moves from 1.94 to 0. When g(X) moves 3 to 2 (X <=90 <= 165) , f(x) moves from 0to -0.966 hence f(x).g(x) moves from 0 to 2 more clearly [0,2). We can say that when g(x) has values 2 to 3 than f(x).g(x) has range (-2, 2).

Same story repeats with negative values. For the rest values when g(x) moves from 0 to 2 CosX moves from -1 to 1 hence f(x).g(x) ranges again in -2 to 2.

Fore clarity, see the graph1 which is depicted for positive g(x) and graph2 which is drawn for negative g(x).

So -2<=CosX(SinX+ (Sin²X+3))<=2

him26.89

sir,

u have written in 2nd last line of inequalities

(-)2

(sin²x + 3)

2

how a root can give negative value?

it should not be (-2) , a negative no.

avinash.sharma

Hello him26.89

Square root of any number always gives a positive and a negative value for example

4 = ± 2

coz

(+2)² = 4 and (-2)² = 4

ia_123

but sir,

4

-2

hence, answer should not be -2

neeraj_agarwal_1990

yes sir ... i think she is correct....
that sign means +ve square root...

isn't it...

pottermania1990

even my math professor told the same thing sir..

i think u shud hav another luk

him26.89

sorry sir,

but i still stick to my point & i am correct.

neeraj_agarwal_1990

same here!!

avinash.sharma

Hello him26.89 and neeraj_agarwal_1990

You both are absolutely right and you should stick to your concepts. But confusion is only due to brief notes. Lets give a look on it

Sin²X+3 = 4

Ö(Sin²X+3 ) = ±Ö (4) <=± 2

it mean the function has values -2 and +2 but when there is a condition of inequality than values move between these -ve and +ve values, which can be proved by preparing a table. I have written the things in short but actually the function

3 <= Sin²X+3 <= 4

or

±Ö3 <= ÖSin²X+3 <= ±Ö4

moves between ±Ö3 and ±Ö4 (When we take sqare root of all the sides). But as the value -2 is less than - Ö3 so I have written directly -2 to +2

Square Roots

The square root is the inverse function of squaring (strictly speaking only for positive numbers, because sign information can be lost)

Principal Root

Every positive number has two square roots, one positive and one negative

Example: 2 is a square root of 4 because 2 ´ 2 = 4, but -2 is also a square root of 4 because (-2) ´ (-2) = 4

To avoid confusion between the two we define the symbol (this symbol is called a radical) to mean the principal or positive square root.

The convention is:

For any positive number x,

is the positive root, and

is the negative root.

If you mean the negative root, use a minus sign in front of the radical.

Example:

_{Sorry for any confusion. All the suggestions are welcomed.}

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