HOTS Zone : Trigonometri

\(\begin{aligned} \sin(\sin(5x-4y)\cdot\sin4(x-y))&=a\\ \sin(5x-4y)\cdot\sin(4x-4y)&=\sin^{-1}a&{\color{red}(1)} \end{aligned}\)

\(\begin{aligned} \cos(\cos(5x-4y)\cdot\cos4(x-y))&=b\\ \cos(5x-4y)\cdot\cos(4x-4y)&=\cos^{-1}b&{\color{red}(2)} \end{aligned}\)

(1) + (2)
\(\begin{aligned} \sin(5x-4y)\cdot\sin(4x-4y)&=\sin^{-1}a\\ \cos(5x-4y)\cdot\cos(4x-4y)&=\cos^{-1}b&{\color{red}+}\\\hline \cos(5x-4y)\cdot\cos(4x-4y)+\sin(5x-4y)\cdot\sin(4x-4y)&=\sin^{-1}a+\cos^{-1}b\\ \cos\left((5x-4y)-(4x-4y)\right)&=\sin^{-1}a+\cos^{-1}b\\ \cos x&=\sin^{-1}a+\cos^{-1}b\\ \cos(\cos x)&=\cos\left(\sin^{-1}a+\cos^{-1}b\right)\\ &=\cos\left(\sin^{-1}a\right)\cdot\cos\left(\cos^{-1}b\right)-\sin\left(\sin^{-1}a\right)\cdot\sin\left(\cos^{-1}b\right)\\ &=\sqrt{1-a^2}\cdotb-a\cdot\sqrt{1-b^2}\\ \end{aligned}\)

\(\begin{aligned} \cos^2 41\degree + \cos^2 43\degree + \cos^2 45\degree + \cos^2 47\degree + \cos^2 49\degree &=\cos^2 41\degree + \cos^2 43\degree + \left(\dfrac12\sqrt2\right)^2 + \sin^2 43\degree + \sin^2 41\degree\\ &=\cos^2 41\degree+ \sin^2 41\degree + \cos^2 43\degree+ \sin^2 43\degree + \dfrac14\cdot2 \\ &=1+1+\dfrac12\\ &=2{,}5 \end{aligned}\)

\(\begin{aligned} \left(\sin\left(15°\right)+\sin\left(75°\right)\right)^2&=\sin^2\left(15°\right)+\sin^2\left(75°\right)+2\sin\left(15°\right)\sin\left(75°\right)\\ &=\sin^2\left(15°\right)+\cos^2\left(15°\right)+2\sin\left(15°\right)\cos\left(15°\right)\\ &=1+sin\left(30°\right)\\ &=1+\dfrac12\\ &=\boxed{\boxed{\dfrac32}} \end{aligned}\)

\(\begin{aligned} \dfrac{3+\cot76°\cot16°}{\cot76°+\cot16°}&=\dfrac{3+\dfrac{\cos76°}{\sin76°}\dfrac{\cos16°}{\sin16°}}{\dfrac{\cos76°}{\sin76°}+\dfrac{\cos16°}{\sin16°}}{\color{red}\cdot\dfrac{\sin76°\sin16°}{\sin76°\sin16°}}\\[4pt] &=\dfrac{3\sin76°\sin16°+\cos76°\cos16°}{\sin16°\cos76°+\cos16°\sin76°}\\[4pt] &=\dfrac{2\sin76°\sin16°+\sin76°\sin16°+\cos76°\cos16°}{\sin\left(16°+76°\right)}\\[4pt] &=\dfrac{\cos\left(76°-16°\right)-\cos\left(76°+16°\right)+\cos\left(76°-16°\right)}{\sin\left(16°+76°\right)}\\[4pt] &=\dfrac{\cos60°-\cos92°+\cos60°}{\sin92°}\\[4pt] &=\dfrac{\dfrac12-\left(1-2\sin^246°\right)+\dfrac12}{2\sin46°\cos46°}\\[4pt] &=\dfrac{2\sin^246°}{2\sin46°\cos46°}\\[4pt] &=\dfrac{\sin46°}{\cos46°}\\ &=\color{blue}\boxed{\boxed{\color{black}\tan46°}} \end{aligned}\)

Perhatikan bahwa $-\dfrac{\pi}2\leq\sin^{-1}\alpha\leq\dfrac{\pi}2$, sehingga $-\dfrac{3\pi}2\leq\sin^{-1}x+\sin^{-1}y+\sin^{-1}z\leq\dfrac{3\pi}2$. Karena $\sin^{-1}x+\sin^{-1}y+\sin^{-1}z=\dfrac{3\pi}2$, maka $\sin^{-1}x=\sin^{-1}y=\sin^{-1}z=\dfrac{\pi}2$. Didapat
$x=y=z=\sin\dfrac{\pi}2=1$

\(\begin{aligned} x^{2023}+y^{2023}+z^{2023}+\dfrac{27}{x^{2023}+y^{2023}+z^{2023}}&=1^{2023}+1^{2023}+1^{2023}+\dfrac{27}{1^{2023}+1^{2023}+1^{2023}}\\[4pt] &=3+\dfrac{27}3\\[4pt] &=3+9\\ &=\color{blue}\boxed{\boxed{\color{black}12}} \end{aligned}\)

\(\begin{aligned} \sqrt{\sin^4x+4\cos^2x}-\sqrt{\cos^4x+4\sin^2x}&=\dfrac13\\[4pt] \sqrt{\sin^4x+4\left(1-\sin^2x\right)}-\sqrt{\cos^4x+4\left(1-\cos^2x\right)}&=\dfrac13\\[4pt] \sqrt{\sin^4x-4\sin^2x+4}-\sqrt{\cos^4x-4\cos^2x+4}&=\dfrac13\\[4pt] \sqrt{\left(\sin^2x-2\right)^2}-\sqrt{\left(\cos^2x-2\right)^2}&=\dfrac13\\[4pt] \left|\sin^2x-2\right|-\left|\cos^2x-2\right|&=\dfrac13 \end{aligned}\)
Perhatikan bahwa sin² x ≤ 1, dan cos² x ≤ 1, maka sin² x − 2 dan cos² x − 2 bernilai negatif, sehingga
\(\begin{aligned} \left|\sin^2x-2\right|-\left|\cos^2x-2\right|&=\dfrac13\\[4pt] 2-\sin^2x-\left(2-\cos^2x\right)&=\dfrac13\\[4pt] 2-\sin^2x-2+\cos^2x&=\dfrac13\\[4pt] \cos^2x-\sin^2x&=\dfrac13\\[4pt] \cos2x&=\dfrac13\\ 6\cos2x&=\color{blue}\boxed{\boxed{\color{black}2}} \end{aligned}\)

\(\begin{aligned} \tan A&=\dfrac{2pq}{p-q}\\[4pt] \dfrac{\sin A}{\cos A}&=\dfrac{2pq}{p-q}\\[4pt] \dfrac{\sqrt{2pq}}{\cos A}&=\dfrac{2pq}{p-q}\\[4pt] \cos A&=p-q \end{aligned}\)

\(\begin{aligned} \sin^2A+\cos^2A&=1\\ \left(\sqrt{2pq}\right)^2+(p-q)^2&=1\\ 2pq+p^2-2pq+q^2&=1\\ p^2+q^2&=\color{blue}\boxed{\boxed{\color{black}1}} \end{aligned}\)

\(\begin{aligned} \sin2a&=2\sin a\cos a\\ \cos a&=\dfrac{\sin 2a}{2\sin a} \end{aligned}\)

\(\begin{aligned} \cos\left(56°\right)\cdot\cos\left(2\cdot56°\right)\cdot\cos\left(2^2\cdot56°\right)\cdots\cos\left(2^{23}\cdot56°\right)&=\dfrac{\sin\left(2\cdot56°\right)}{2\sin\left(56°\right)}\cdot\dfrac{\sin\left(2^2\cdot56°\right)}{2\sin\left(2\cdot56°\right)}\cdot\dfrac{\sin\left(2^3\cdot56°\right)}{2\sin\left(2^2\cdot56°\right)}\cdots\dfrac{\sin\left(2^{24}\cdot56°\right)}{2\sin\left(2^{23}\cdot56°\right)}\\[4pt] &=\dfrac{\sin\left(2^{24}\cdot56°\right)}{2^{24}\sin\left(56°\right)} \end{aligned}\)
Kita buktikan bahwa 2²⁴⋅56° = 56° + k⋅360°, atau 2²⁴⋅56 = 56 mod 360.
\(\begin{aligned} 2^{24}\cdot56\mod360&\equiv8\left(2^{24}\cdot7\mod45\right)\mod360\\ &\equiv8\left(7\mod45\right)\mod360&\ {\color{red}\phi(45)=24}\\ &\equiv56\mod360 \end{aligned}\)

\(\begin{aligned} \dfrac{\sin\left(2^{24}\cdot56°\right)}{2^{24}\sin\left(56°\right)}&=\dfrac{\sin\left(56°\right)}{2^{24}\sin\left(56°\right)}\\[4pt] &=\dfrac1{2^{24}}\\ &=\color{blue}\boxed{\boxed{\color{black}2^{-24}}} \end{aligned}\)