Question

If x and y are the maximum and minimum values of sin $\theta$ + cos $\theta$, find the value of $x^2$ + $y^2$

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Answer 1

If you know that the maximum and minimum value of asin$\theta$ + bcos$\theta$ are $\sqrt{a^2+b^2}$, and -$\sqrt{a^2+b^2}$,you can directly use this result to arrive at the answer.

We will try to arrive at the answer similar to the derivation of the above result.

Let E = asin$\theta$ + bcos$\theta$, If we can express it as k sin (A + B), we can say the minimum and maximum values are -k and +k, where k is a +ve number.

we will expand ksin (A+B) and campare it with asin$\theta$ + bsin$\theta$.

ksin(A+B) = asin$\theta$ + bcos$\theta$

k[sinAcosB + cosAsinB) = asin$\theta$ + bcos$\theta$

If we take A = $\theta$, then we get k cos B = a and Ksin B = b ___________(1)

$\Rightarrow$ cosB = $\frac{a}{k}$ and sinB = $\frac{b}{k}$

$co{s}^{2}B$ + $si{n}^{2}B$ = $\frac{a^2+b^2}{k^2}$ = 1

$\Rightarrow$ k = $\sqrt{a^2+b^2}$

From (1) we get = cosB = $\frac{a}{\sqrt{a^2+b^2}}$

sinB = $\frac{b}{\sqrt{a^2+b^2}}$

So we will divide and multiply by $\sqrt{a^2+b^2}$ in asin$\theta$ + bcos$\theta$.

$\Rightarrow$ $\sqrt{a^2+b^2}$ $(\frac{a}{\sqrt{a^2+b^2}}sin\theta +\frac{b}{\sqrt{a^2+b^2}}cos\theta )$

$\Rightarrow$ asin$\theta$ + bcos$\theta$ = $\sqrt{a^2+b^2}$ (sin$\theta$x cosB + cos$\theta$sinB)
= $\sqrt{a^2+b^2}$ sin($\theta$ + B)

$\to$ Maximum and minimum values are

$\sqrt{a^2+b^2}$ and - $\sqrt{a^2+b^2}$

In our question a = 1 , b =1

$\Rightarrow$ The minimum and maximum values are $-\sqrt{2}$ and $\sqrt{2}$

$\Rightarrow$ $x^2$ + $y^2$ = 2 + 2 = 4