Jika
[tex]\begin{aligned}a&=\frac{1+\sin 88^\circ+\cos 88^\circ}{1+\sin 88^\circ-\cos 88^\circ}\\b&=\left ( 1+\tan 44^\circ\tan 22^\circ \right )\sin 44^\circ\\c&=\sin 44^\circ\cos 44^\circ\end{aligned}[/tex]
tentukan nilai dari [tex]\displaystyle (a+b)c[/tex]
[tex]\begin{aligned}a&=\frac{1+\sin 88^\circ+\cos 88^\circ}{1+\sin 88^\circ-\cos 88^\circ}\\b&=\left ( 1+\tan 44^\circ\tan 22^\circ \right )\sin 44^\circ\\c&=\sin 44^\circ\cos 44^\circ\end{aligned}[/tex]
tentukan nilai dari [tex]\displaystyle (a+b)c[/tex]
Verified answer
[tex]\begin{aligned}&\textsf{Nilai dari }(a+b)c {\sf\ adalah\ }\:\boxed{\bf1}\:.\end{aligned}[/tex]
Penjelasan dengan langkah-langkah:
Trigonometri
Diberikan nilai-nilai:
[tex]\left\{\begin{aligned}a&=\frac{1+\sin 88^\circ+\cos 88^\circ}{1+\sin 88^\circ-\cos 88^\circ}\\b&=\left (1+\tan 44^\circ\tan 22^\circ \right )\sin 44^\circ\\c&=\sin 44^\circ\cos 44^\circ\end{aligned}\right.[/tex]
Kita akan menentukan nilai dari [tex](a+b)c[/tex].
Pertama-tama,
[tex]\begin{aligned}&\frac{1+\sin x+\cos x}{1+\sin x-\cos x}\\&{=\ }\frac{\sin x(1+\sin x+\cos x)}{\sin x(1+\sin x-\cos x)}\\&{=\ }\frac{\sin x(1+\cos x)+\sin^2x}{\sin x(1+\sin x-\cos x)}\\&{=\ }\frac{\sin x(1+\cos x)+1-\cos^2x}{\sin x(1+\sin x-\cos x)}\\&{=\ }\frac{\sin x(1+\cos x)+(1-\cos x)(1+\cos x)}{\sin x(1+\sin x-\cos x)}\\&{=\ }\frac{(1+\cos x)\cancel{(1+\sin x-\cos x)}}{\sin x\cancel{(1+\sin x-\cos x)}}\end{aligned}[/tex]
[tex]\begin{aligned}&{=\ }\frac{1+\cos x}{\sin x}\\&{=\ }\frac{\cancel{2}\cos^2\left(\dfrac{x}{2}\right)}{\cancel{2}\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right)}\\&{=\ }\frac{\cos\left(\dfrac{x}{2}\right)}{\sin\left(\dfrac{x}{2}\right)}\\&{=\ }\cot\left(\dfrac{x}{2}\right)\end{aligned}[/tex]
Maka:
[tex]\begin{aligned}a&=\frac{1+\sin 88^\circ+\cos 88^\circ}{1+\sin 88^\circ-\cos 88^\circ}\\&=\cot\left(\frac{88^\circ}{2}\right)\\a&=\cot44^\circ\end{aligned}[/tex]
Lalu,
[tex]\begin{aligned}&(1+\tan 2x\tan x)\sin 2x\\&{=\ }\left(1+\frac{\sin2x\sin x}{\cos2x\cos x}\right)\sin 2x\\&{=\ }\left(\frac{\cos2x\cos x+\sin2x\sin x}{\cos2x\cos x}\right)\sin 2x\\&{=\ }\left(\frac{\cos(2x-x)}{\cos2x\cos x}\right)\sin 2x\\&{=\ }\left(\frac{\cancel{\cos x}}{\cos2x\cancel{\cos x}}\right)\sin 2x\\&{=\ }\tan2x\end{aligned}[/tex]
Maka,
[tex]\begin{aligned}b&=\left(1+\tan 44^\circ\tan 22^\circ\right)\sin 44^\circ\\&\quad\rightarrow\ x=22^\circ\ \Rightarrow\ 2x=44^\circ\\b&=\tan44^\circ\end{aligned}[/tex]
Kemudian,
[tex]\begin{aligned}&(\cot x+\tan x)\sin x\cos x\\&{=\ }(\cot x\sin x)\cos x+(\tan x\cos x)\sin x\\&{=\ }\cos^2x+\sin^2x\\&{=\ }1\end{aligned}[/tex]
Oleh karena itu,
[tex]\begin{aligned}(a+b)c&=(\cot44^\circ+\tan44^\circ)\sin44^\circ\cos44^\circ\\\because\ &(\cot x+\tan x)\sin x\cos x=1\\\therefore\ (a+b)c&=\boxed{\bf1}\quad\blacksquare\\\end{aligned}[/tex]