Question

(cosx+i sinx)^6÷(sinx+i cosx)^6​

Answer 1

Answer:

sin(x) and cos(x) both ranges from -1 to +1 ….but since sin(x) is increasing for 0<=x<=π/2; and cos(x) is decreasing at the same time for 0<=x<=π/2; ….and for π/2<=x<=π sin(x) is decreasing and cos (x) is increasing..... and for other x likewise… ie... we can't say directly that.... Both ranges from -1 to +1 hence their 6th power also ranges from -1 to +1 …hence their sum is ranges from -2 to +2 …. So what we do...

Let's try to simplify the question....

(sin(x))^6 +(cos(x))^6 == ((sin(x))^2)^3 + ((cos(x))^2)^3

According to formula. a^3 +b^3 = (a+b)*(a^2 + b^2 - a*b)

== ( (sin(x))^2 + (cos(x))^2 ) * ( (sin(x))^4 + (cos(x))^4 - (sin(x))^2 * (cos(x))^2 )

As (sin(x))^2 + (cos(x))^2 = 1

And a^2 + b ^2 = (a+b)^2 - 2*a*b ;

== 1* ( ( (sin(x))^2 + (cos (x))^2) )^2 - 3 * (sin(x))^2 * (cos(x))^2 )

== 1* ( (1)^2 - (3/4) * (sin(2*x))^2 )

== ( 1 - (3/4) * (sin(2*x))^2 )

Since -1<=sin(2*x) <=1 for all x € R;

Hence -1<=(sin(2*x))^2 <=1 for all x € R;

Hence (-3/4)<= - (3/4)* (sin(2*x))^2 <= (3/4) ;

Hence 1 + (-3/4) <= ( 1 - (3/4) * (sin(2*x))^2 ) <= 1 + (3/4) ;

Hence (1/4) <= F(x) <= (7/4) .