Prove that: `forall x\inR` : `:|acosx+bsinx|\leqsqrt(a^2+b^2)`

a. Prove that: `forall x\inR` : `:|acosx+bsinx|\leqsqrt(a^2+b^2)`

b. Find the maximum and minimum of `f(x)=20cosx+21sinx+27`

Solution

a. Prove `|acosx+bsinx|\leqsqrt(a^2+b^2)` `forall x\inR`

Assuming we have: 

`acosx+bsinx=sqrt(a^2+b^2)(a/sqrt(a^2+b^2)cosx+b/sqrt(a^2+b^2)sinx)`

Let  `sin\alpha=a/sqrt(a^2+b^2)` which's `-1\leqsin\alpha\leq1`

Then, `cos\alpha=sqrt(1-sin^2\alpha)` because `sin^2\alpha+cos^2\alpha=1`

            `cos\alpha=sqrt(1-(a/sqrt(a^2+b^2))^2)=sqrt(1-a^2/(a^2+b^2))=b/(sqrt(a^2+b^2)`

Therefore: `acosx+bsinx=sqrt(a^2+b^2)(sin\alphacosx+cos\alphasinx)`

By using formula `sinacosb+sinbcosa=sin(a+b)`

Then `acosx+bsinx=sqrt(a^2+b^2)sin(x+\alpha)`

         `|acosx+bsinx|=sqrt(a^2+b^2)|sin(x+\alpha)|`

As we knew: `forall x;\alpha\inR` `-1\leqsin(x-+\alpha)\leq1`

Then, `|sin(x+\alpha)|\leq1`

          `sqrt(a^2+b^2)|sin(x+\alpha)|\leqsqrt(a^2+b^2)`

Hence, `|acosx+bsinx|\leqsqrt(a^2+b^2)`

b. Find the max and min of `f(x)=20cosx+21sinx+27`

        `f(x)=27+20cosx+21sinx`

              `=27+29(20/29cosx+21/29sinx)`

              `=27+29(sin\alphacosx+cos\alphasinx)` which `\alpha=arcsin(20/29)`

             `=27+29sin(x+\alpha)`

As we proved above: `-29\leq29sin(x+\alpha)\leq29`

Then, `27-29\leq27+29sin(x+\alpha)\leq27+29`

           `-2\leqf(x)\leq56`

Thus, Max`f(x)=56` and Min`f(x)=-2`

Solution by: Thin Sokkean