Question

a=(sin theta- cos theta)^4,b=sin^6 theta+cos^6 theta and c=(sin theta +cos theta)^2, then the value of √3a+4b+6c lies between

Answer 1

√3a+4b+6c  = √13  when a = (Sinθ - Cosθ)⁴   , b = Sin⁶θ + Cos⁶θ & c = (Sinθ + Cosθ)²

Step-by-step explanation:

a = (Sinθ - Cosθ)⁴  

=>a =  ((Sinθ - Cosθ)²)²

=> a = (Sin²θ + Cos²θ - 2SinθCosθ)²

=> a = (1 - 2SinθCosθ)²

=> a =  1  + 4Sin²θCos²θ - 4SinθCosθ

b =  Sin⁶θ + Cos⁶θ

using x³ + y³ = (x + y)³  - 3xy(x + y)

x = Sin²θ  & y  = Cos²θ

=> b = (Sin²θ + Cos²θ)³  - 3(Sin²θ * Cos²θ)(Sin²θ + Cos²θ)

=> b = 1 -  3Sin²θCos²θ

c = (Sinθ + Cosθ)²

=> c = 1 + 2SinθCosθ

3a + 4b + 6c

= 3  + 12Sin²θCos²θ - 12SinθCosθ + 4 - 12Sin²θCos²θ  + 6 + 12SinθCosθ

= 13

√3a+4b+6c  = √13

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Answer 2

So the value of  $\sqrt{3a+4b+6c}$ is $\sqrt{13}$

Step-by-step explanation:

Given;

$a=(sin\theta-cos\theta)^{4}$ , $b=sin^{6}\theta +cos^{6}\theta$ and $c=(sin\theta+cos\theta)^{2}$

From,

  $a=(sin\theta-cos\theta)^{4}$

⇒$a=((sin\theta-cos\theta)^{2})^{2}$

⇒$a=(sin^{2} \theta+cos^{2} \theta-2sin\theta cos\theta)^{2}$    (∵$(a-b)^{2} =a^{2} -2ab+b^{2}$)

⇒$a=(1-2sin\theta cos\theta)^{2}$   (∵$sin^{2} \theta+cos^{2} \theta=1$)

⇒$a=1-4sin\theta cos\theta+4sin^{2} \theta cos^{2} \theta$

And,

  $b=sin^{6}\theta +cos^{6}\theta$

⇒$b=(sin^{2} \theta)^{3} +(cos^{2} \theta)^{3}$

⇒$b=(sin^{2} \theta+cos^{2} \theta)^{3}-3sin^{2} \theta cos^{2} \theta(sin^{2} \theta +cos^{2} \theta)$  (using $x^{3} + y^{3} = (x + y)^{3} - 3xy(x + y)$ )

⇒$b=1-3sin^{2} \theta cos^{2} \theta$

Also,

  $c=(sin\theta+cos\theta)^{2}$

⇒$c=sin^{2} \theta+cos^{2} \theta+2sin\theta cos\theta$

⇒$c=1+2sin\theta cos\theta$

Now,

  $\sqrt{3a+4b+6c}$

Plug all the values in above equation;

$\sqrt{3(1-4sin\theta cos\theta+4sin^{2} \theta cos^{2} \theta)+4(1-3sin^{2} \theta cos^{2} \theta)+6(1+2sin\theta cos\theta)}$

$=\sqrt{(3-12sin\theta cos\theta+12sin^{2} \theta cos^{2} \theta+4-12sin^{2} \theta cos^{2} \theta+6+12sin\theta cos\theta)}$

$=\sqrt{13}$

Therefore the value of  $\sqrt{3a+4b+6c}$ is $\sqrt{13}$