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Question Number 106958 by bemath last updated on 08/Aug/20

@ b e m a t h @

(1) G i v e n {\begin{cases} x = \sin α + \sin β \\ y = \cos α + \cos β \end{cases}

m a x i m u m v a l u e o f x^{2} + y^{2} w h e n α = \dots

(2) f i n d s o l u t i o n s e t t h e e q u a t i o n

\sin^{4} x + \sin^{4} (x + \frac{π}{4}) = \frac{1}{4} w h e r e x \in [0, 2 π]

Answered by mr W last updated on 08/Aug/20

(1)

x^{2} + y^{2} = \sin^{2} α + \sin^{2} β + 2 \sin α \sin β +

\cos^{2} α + \cos^{2} β + 2 \cos α \cos β

= 2 + 2 \cos (α - β)

\Rightarrow {(x^{2} + y^{2})}_{m a x} = 2 + 2 = 4

\Rightarrow {(x^{2} + y^{2})}_{m i n} = 2 - 2 = 0

Commented by bemath last updated on 08/Aug/20

t h a n k y o u m r W

Commented by bemath last updated on 08/Aug/20

{(x^{2} + y^{2})}_{m a x} w h e r e \cos (α - β) = 1

\to \cos (α - β) = \cos 0 °

\to α - β = 0 ° + k . 2 π, \Rightarrow α = β + k . 2 π

Answered by mr W last updated on 08/Aug/20

(2)

\sin^{4} x + \frac{1}{4} {(\sin x + \cos x)}^{4} = \frac{1}{4}

\sin^{4} x + \frac{1}{4} {(1 + 2 \sin x \cos x)}^{2} = \frac{1}{4}

\sin^{4} x + \frac{1}{4} (1 + 4 \sin x \cos x + 4 \sin^{2} x \cos^{2} x) = \frac{1}{4}

\sin^{4} x + \frac{1}{4} + \sin x \cos x + \sin^{2} x \cos^{2} x = \frac{1}{4}

\sin^{2} x + \sin x \cos x = 0

\sin x (\sin x + \cos x) = 0

\sqrt{2} \sin x \sin (x + \frac{π}{4}) = 0

\Rightarrow \sin x = 0 \Rightarrow x = k π

\Rightarrow \sin (x + \frac{π}{4}) = 0 \Rightarrow x = k π - \frac{π}{4}

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