Question

TRIGONOMETRI
Jika P= 2 sin @ dan R= √2 cos @,
maka ½P²+ R² = ...... ?
A. ½
B. 1
C. √2
D 2
E. 2√2

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Answer 1

Verified answer

Jawaban:

D. 2

Penjelasan dengan langkah-langkah:

[tex]p = 2 \sin( \alpha ) \\ r = \sqrt{2} \cos( \alpha ) \\ \frac{1}{2} p^{2} + {r}^{2} \\ = \frac{1}{2} {(2 \sin( \alpha )) }^{2} + (\sqrt{2} \cos( \alpha ) ) {}^{2} \\ = \frac{1}{2} 4 \sin^{2} ( \alpha ) + 2 {cos}^{2} ( \alpha ) \\ = 2 {sin}^{2} ( \alpha ) + 2 \cos {}^{2} ( \alpha ) \\ = 2(1) \\ = 2[/tex]

Answer 2

Penjelasan dengan langkah-langkah:

[tex]\displaystyle \rm \large \boxed{\color{orange}{ p = 2 \sin \: \alpha }}[/tex]

[tex]\displaystyle \rm \large \boxed{ \color{orange}{ r= \sqrt{2} \cos \: \alpha }}[/tex]

Maka:

[tex]\displaystyle \rm \large \color{orange}{ \frac{ {p}^{2} }{2} + {r}^{2} }[/tex]

[tex]\displaystyle \rm \large \color{orange}{ = \frac{ {(2 \sin \alpha ) }^{2} }{2} + { (\sqrt{2} \cos \alpha) }^{2} }[/tex]

[tex]\displaystyle \rm \large \color{orange}{ = \frac{ {\bcancel{4}\sin}^{2} \alpha }{\bcancel{2}} + {2\cos}^{2}\alpha}[/tex]

[tex]\displaystyle \rm \large \color{orange}{ ={2 \sin}^{2} \alpha + {2\cos}^{2}\alpha}[/tex]

[tex]\displaystyle \rm \large \color{orange}{ = 2( { \sin}^{2}a + { \cos}^{2}a) }[/tex]

[tex] \displaystyle \rm \large \color{orange}{ = 2(1) }[/tex]

[tex] = \displaystyle \rm \large \color{orange}{ 2}[/tex]

[tex]\boxed{ \colorbox{red}{ \sf{ \color{yellow}{ \: Ω RedAsta Ω \: }}}}[/tex]