5.5 Rumus bagi sin (A ± B), kos (A ± B), tan (A ± B), sin 2A, kos 2A, tan 2A

5.5 Rumus bagi sin (A ± B), kos (A ± B), tan (A ± B), sin 2A, kos 2A, tan 2A
Rumus penambahan
\(\boxed{\begin{aligned} &\sin A=2 \sin \frac{A}{2} k \operatorname{sos} \frac{A}{2}\\ &k o s A=\sin ^{2} \frac{A}{2}-\operatorname{kos}^{2} \frac{A}{2}\\ &\operatorname{kos} A=2 \operatorname{kos}^{2} \frac{A}{2}-1\\ &k o s A=1-2 k o s^{2} \frac{A}{2}\\ &\text { – } \tan A=\frac{2 \tan \frac{A}{2}}{1-\tan ^{2} \frac{A}{2}} \end{aligned}}\)
5.5.1 Pembuktian Identiti Trigonometri yang Melibatkan Sudut Majmuk dan Sudut Berganda
Contoh 1:
Buktikan setiap identity trigonometri yang berikut.
\(\begin{array}{l}\text{(a) }\frac{sin{(A+B)}-sin{(A-B)}}{ko\mathrm sAkosB}=2\tan B\\\text{(b) }\frac{kos{(A+B)}}{\sin AkosB}=kotA-\tan B\\\text{(c) }\tan{(A+45^o)}=\frac{\sin A+kosA}{ko\mathrm sA-\sin A}\end{array}\)  
penyelesaian:
(a)
\(\begin{array}{l}(Sebelah\text{ }Kiri)\\=\frac{sin{(A+B)}-sin{(A-B)}}{ko\mathrm sAkosB}\\=\frac{{(\sin AkosB+kosA\sin B)}-{(\sin AkosB-kosA\sin B)}}{ko\mathrm sAkosB}\\=\frac{2\cancel{kosA}\sin B}{\cancel{kosA}kosB}\\=\frac{2\sin B}{kosB}\\=2\tan B=(Sebelah\text{ }Kanan)\end{array}\)
(b)
\(\begin{array}{l}(Sebelah\text{ }Kiri)\\=\frac{kos{(A+B)}}{\sin AkosB}\\=\frac{kosAkosB-\sin A\sin B}{\sin AkosB}\\=\frac{kosA\cancel{kosB}}{\sin A\cancel{kosB}}-\frac{\cancel{\sin A}\sin B}{\cancel{\sin A}kosB}\\=\frac{kosA}{\sin A}-\frac{\sin B}{kosB}\\=kotA-\tan B\\=(Sebelah\text{ }Kanan)\end{array}\)
(c)
\(\begin{array}{l}(Sebelah\text{ }Kiri)\\=\tan{(A+45^o)}\\=\frac{tanA+\tan45^o}{1-tanA\tan45^o}\\=\frac{tanA+1}{1-tanA}\leftarrow\boxed{\tan45^o=1}\\=\frac{\frac{\sin A}{kosA}+1}{1-\frac{\sin A}{kosA}}\\=\frac{\sin A+kosA}{\cancel{kosA}}\times\frac{\cancel{kosA}}{kosA-\sin A}\\=\frac{\sin A+kosA}{kosA-\sin A}\\=(Sebelah\text{ }Kanan)\end{array}\)

Leave a Comment